A -

 Wilbur and Swimming Pool

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

   

Description

After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the area of the rectangle must be positive. Wilbur had all four vertices of the planned pool written on a paper, until his friend came along and erased some of the vertices.

Now Wilbur is wondering, if the remaining n vertices of the initial rectangle give enough information to restore the area of the planned swimming pool.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 4) — the number of vertices that were not erased by Wilbur's friend.

Each of the following n lines contains two integers x_i and y_i ( - 1000 ≤ x_i, y_i ≤ 1000) —the coordinates of the i-th vertex that remains. Vertices are given in an arbitrary order.

It's guaranteed that these points are distinct vertices of some rectangle, that has positive area and which sides are parallel to the coordinate axes.

Output

Print the area of the initial rectangle if it could be uniquely determined by the points remaining. Otherwise, print  - 1.

Sample Input

Input

2 0 0 1 1

Output

1

Input

1 1 1

Output

-1

Hint

In the first sample, two opposite corners of the initial rectangle are given, and that gives enough information to say that the rectangle is actually a unit square.

In the second sample there is only one vertex left and this is definitely not enough to uniquely define the area.

题解：水题，给一系列点，问是否能构成唯一矩形；

代码：

A#include
        
         #include
         
          #include
          
           #include
           
            using namespace std;#define PI(x) printf("%d",x)#define SI(x) scanf("%d",&x)#define P_(x) printf(" ")const int INF=0x3f3f3f3f;typedef long long LL;struct Dot{    int x,y;};Dot a[5];int main(){    int N;    while(~scanf("%d",&N)){        int lx,rx,ly,ry;        for(int i=0;i
           
          
         
        

B -

 Tricky Sum

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

   

Description

In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum.

For example, for n = 4 the sum is equal to  - 1 - 2 + 3 - 4 =  - 4, because 1, 2 and 4 are 2⁰, 2¹ and 2² respectively.

Calculate the answer for t values of n.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of values of n to be processed.

Each of next t lines contains a single integer n (1 ≤ n ≤ 10⁹).

Output

Print the requested sum for each of t integers n given in the input.

Sample Input

Input

2 4 1000000000

Output

-4 499999998352516354

Hint

The answer for the first sample is explained in the statement.

 题解：求- 1 - 2 + 3 - 4.....的等式结果，其中 2⁰, 2¹ and 2²....前是负号；

代码：

B#include
           
            #include
            
             #include
             
              #include
              
               using namespace std;#define PI(x) printf("%d",x)#define SI(x) scanf("%d",&x)#define P_(x) printf(" ")const int INF=0x3f3f3f3f;typedef long long LL;int main(){    LL n;    int T;    SI(T);    while(T--){        LL temp=0,i=0;        scanf("%lld",&n);        while(pow(2,i)<=n){            temp+=pow(2,i);            i++;        }        printf("%lld\n",n*(n+1)/2-2*temp);    }    return 0;}
              
             
            
           

D -

 Interview

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

   

Description

Blake is a CEO of a large company called "Blake Technologies". He loves his company very much and he thinks that his company should be the best. That is why every candidate needs to pass through the interview that consists of the following problem.

We define function f(x, l, r) as a bitwise OR of integers x_l, x_l + 1, ..., x_r, where x_i is the i-th element of the array x. You are given two arrays a and b of length n. You need to determine the maximum value of sum f(a, l, r) + f(b, l, r) among all possible 1 ≤ l ≤ r ≤ n.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the length of the arrays.

The second line contains n integers a_i (0 ≤ a_i ≤ 10⁹).

The third line contains n integers b_i (0 ≤ b_i ≤ 10⁹).

Output

Print a single integer — the maximum value of sum f(a, l, r) + f(b, l, r) among all possible 1 ≤ l ≤ r ≤ n.

Sample Input

Input

5 1 2 4 3 2 2 3 3 12 1

Output

22

Input

10 13 2 7 11 8 4 9 8 5 1 5 7 18 9 2 3 0 11 8 6

Output

46

Hint

Bitwise OR of two non-negative integers a and b is the number c = aORb, such that each of its digits in binary notation is 1 if and only if at least one of a or b have 1 in the corresponding position in binary notation.

In the first sample, one of the optimal answers is l = 2 and r = 4, because f(a, 2, 4) + f(b, 2, 4) = (2 OR 4 OR 3) + (3 OR 3 OR12) = 7 + 15 = 22. Other ways to get maximum value is to choose l = 1 and r = 4, l = 1 and r = 5, l = 2 and r = 4, l = 2 and r = 5,l = 3 and r = 4, or l = 3 and r = 5.

In the second sample, the maximum value is obtained for l = 1 and r = 9.

题解：水，求两个数组的或值和最大；由于是或，必然是所有的或最大；

代码：

D#include
              
               #include
               
                #include
                
                 #include
                 
                  using namespace std;#define PI(x) printf("%d",x)#define SI(x) scanf("%d",&x)#define P_(x) printf(" ")const int INF=0x3f3f3f3f;const int MAXN=1010;typedef long long LL;LL a[MAXN],b[MAXN],c[MAXN];int main(){    int n;    while(~scanf("%d",&n)){        LL x1=0,x2=0;        for(int i=0;i
                 
                
               
              

F -

 Infinite Sequence

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

   

Description

Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).

Find the number on the n-th position of the sequence.

Input

The only line contains integer n (1 ≤ n ≤ 10¹⁴) — the position of the number to find.

Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Output

Print the element in the n-th position of the sequence (the elements are numerated from one).

Sample Input

Input

3

Output

2

Input

5

Output

2

Input

10

Output

4

Input

55

Output

10

Input

56

Output

1 题解：给位置，问对应值；看着数那么大，果断二分了，然而错了。。。不知道为啥。。。然后暴力下过了； 代码：

F#include
                  
                   #include
                   
                    #include
                    
                     #include
                     
                      using namespace std;#define PI(x) printf("%d",x)#define SI(x) scanf("%d",&x)#define P_(x) printf(" ")const int INF=0x3f3f3f3f;typedef long long LL;LL n;/*LL erfen(LL l,LL r){    LL mid;    while(l<=r){        mid=(l+r)/2;        if(mid*(mid+1)/2>=n)r=mid-1;        else l=mid+1;    }    return r;}*/int main(){    while(~scanf("%lld",&n)){        LL ans;        /*ans=erfen(1,n);    //    printf("%lld\n",ans);        while(ans*(ans+1)/2>=n)ans--;        */        LL i;        for(i=1;;i++){            if(n-i<=0)break;            n-=i;        }        printf("%lld\n",n);    }    return 0;}
                     
                    
                   
                  

G -

 Slime Combining

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

   

Description

Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.

You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.

You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right.

Input

The first line of the input will contain a single integer, n (1 ≤ n ≤ 100 000).

Output

Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.

Sample Input

Input

1

Output

1

Input

2

Output

2

Input

3

Output

2 1

Input

8

Output

4

Hint

In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.

In the second sample, we perform the following steps:

Initially we place a single slime in a row by itself. Thus, row is initially 1.

Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.

In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.

In the last sample, the steps look as follows:

1. 1
2. 2
3. 2 1
4. 3
5. 3 1
6. 3 2
7. 3 2 1
8. 4

题解：

就跟2048那个游戏一样，不过这个更简单，是一维的。。。

代码：

G#include
                    
                     #include
                     
                      #include
                      
                       #include
                       
                        using namespace std;#define PI(x) printf("%d",x)#define SI(x) scanf("%d",&x)#define P_(x) printf(" ")const int INF=0x3f3f3f3f;typedef long long LL;const int MAXN=1000100;LL ans[MAXN];int main(){    int n;    while(~scanf("%d",&n)){        int tp=0;        while(n--){            ans[tp]=1;            tp++;            while(tp>1&&ans[tp-1]==ans[tp-2]){                ans[tp-2]=ans[tp-2]+1;                tp--;            }        }        for(int i=0;i
                       
                      
                     
                    

H -

 Baby Ming and phone number

Time Limit:1500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

   

Description

Baby Ming collected lots of cell phone numbers, and he wants to sell them for money.

 

He thinks normal number can be sold for $b$ yuan, while number with following features can be sold for $a$ yuan. 

1.The last five numbers are the same. (such as 123-4567-7777) 

2.The last five numbers are successive increasing or decreasing, and the diffidence between two adjacent digits is $1$. (such as 188-0002-3456) 

3.The last eight numbers are a date number, the date of which is between Jar 1st, 1980 and Dec 31th, 2016. (such as 188-1888-0809,means August ninth,1888) 

Baby Ming wants to know how much he can earn if he sells all the numbers. 

 

Input

In the first line contains a single positive integer $T$, indicating number of test case.

 

In the second line there is a positive integer $n$, which means how many numbers Baby Ming has.(no two same phone number) 

In the third line there are $2$ positive integers $a, b$, which means two kinds of phone number can sell $a$ yuan and $b$ yuan. 

In the next $n$ lines there are $n$ cell phone numbers.（|phone number|==11, the first number can’t be 0) 

$1 \leq T \leq 30, b < 1000, 0 < a, n \leq 100,000$ 

 

Output

How much Baby Nero can earn.

 

Sample Input

1 5 100000 1000 12319990212 11111111111 22222223456 10022221111 32165491212

 

Sample Output

302000

题解：

卖电话号，定义一定规则的电话号是好的，好的a元钱，普通的b元；问能卖多少钱；注意细节就好了；

代码：

H#include
                     
                      #include
                      
                       #include
                       
                        #include
                        
                         using namespace std;#define PI(x) printf("%d",x)#define SI(x) scanf("%d",&x)#define P_(x) printf(" ")const int INF=0x3f3f3f3f;typedef long long LL;int num[12];int mon[13]={
                   0,31,28,31,30,31,30,31,31,30,31,30,31};bool rn(int y){    if(y%400==0||(y%100!=0&&y%4==0))return true;    return false;}bool js(){//    for(int i=1;i<=11;i++)printf("%d ",num[i]);puts("");    int temp=1;    for(int i=7;i<11;i++){        if(num[i]!=num[i+1])temp=0;    }    if(temp)return true;//    puts("1");        temp=1;    for(int i=7;i<11;i++){        if(num[i]+1!=num[i+1])temp=0;    //    printf("%d\n",temp);    }    if(temp)return true;//    puts("2");        temp=1;    for(int i=7;i<11;i++){        if(num[i]-1!=num[i+1])temp=0;    }    if(temp)return true;        int year=0,mm=0,rr=0;    for(int i=4;i<4+4;i++)year=year*10+num[i];    for(int i=8;i<8+2;i++)mm=mm*10+num[i];    for(int i=10;i<10+2;i++)rr=rr*10+num[i];    if(!(year>=1980&&year<=2016))return false;    if(rr==0||rr>31||mm==0||mm>12)return false;    if(!rn(year)&&mm==2&&rr>=29)return false;    if(mm==2&&rr>29)return false;    if(rn(year)&&mm==2&&rr==29)return true;    if(rr>mon[mm])return false;//    puts("3");    return true;}int main(){    int T,n,a,b;    SI(T);    char s[15];    while(T--){        SI(n);        scanf("%d%d",&a,&b);        LL ans=0;        for(int i=0;i
                        
                       
                      
                     

I -

 Divisibility

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

   

Description

Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x thata ≤ x ≤ b and x is divisible by k.

Input

The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 10¹⁸; - 10¹⁸ ≤ a ≤ b ≤ 10¹⁸).

Output

Print the required number.

Sample Input

Input

1 1 10

Output

10

Input

2 -4 4

Output

5 题解：问a,b之间有多少个数能被k整除，如果都大于0处理a；如果都小于0，处理b，这点让我wa了两次；一个大0一个小0，则加上0； 代码：

I#include
                         
                          #include
                          
                           #include
                           
                            #include
                            
                             using namespace std;#define PI(x) printf("%d",x)#define SI(x) scanf("%d",&x)#define P_(x) printf(" ")const int INF=0x3f3f3f3f;typedef __int64 LL;int main(){    LL k,a,b;    while(~scanf("%I64d%I64d%I64d",&k,&a,&b)){        LL x1=a/k;        LL x2=b/k;        LL ans=x2-x1;        if(a==0&&b==0){            puts("1");continue;        }        if(a>=0&&b>=0&&a%k==0)ans++;        else if(a<=0&&b<=0&&b%k==0)ans++;        else if(a<=0&&b>=0)ans++;        printf("%I64d\n",ans);    }    return 0;}