这一场PP说很简单，我就闲来无事敲一敲好了，反正也考完了（雾，不过只剩数据结构和c++

A. Okabe and Future Gadget Laboratory

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an n by nsquare grid of integers. A good lab is defined as a lab in which every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every x, y such that 1 ≤ x, y ≤ n and ax, y ≠ 1, there should exist two indices s and t so that ax, y = ax, s + at, y, where ai, j denotes the integer in i-th row and j-th column.

Help Okabe determine whether a given lab is good!

Input

The first line of input contains the integer n (1 ≤ n ≤ 50) — the size of the lab.

The next n lines contain n space-separated integers denoting a row of the grid. The j-th integer in the i-th row is ai, j (1 ≤ ai, j ≤ 105).

Output

Print "Yes" if the given lab is good and "No" otherwise.

You can output each letter in upper or lower case.

Examples

input

3 1 1 2 2 3 1 6 4 1

output

Yes

input

3 1 5 2 1 1 1 1 2 3

output

No

Note

In the first sample test, the 6 in the bottom left corner is valid because it is the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this table, so the answer is "Yes".

In the second sample test, the 5 cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".

 这个题的意思就是给你一个矩阵，假如不是1的话让你找一个所在行的数和所在列的数是不是和他想等，正确yes，否则no。暴力就可以的

#include 
          
           int a[55][55];int n;int solve(int x,int y) {    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)            if(a[x][i]+a[j][y]==a[x][y]) return 1;    return 0;}int main() {    scanf("%d",&n);    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)            scanf("%d",a[i]+j);    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)            if(a[i][j]!=1)                if(solve(i,j)==0)                    return 0*printf("No");    return 0*printf("Yes");}
          

 

 

B. Okabe and Banana Trees

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees.

Consider the point (x, y) in the 2D plane such that x and y are integers and 0 ≤ x, y. There is a tree in such a point, and it has x + ybananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation . Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point.

Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely.

Okabe is sure that the answer does not exceed 1018. You can trust him.

Input

The first line of input contains two space-separated integers m and b (1 ≤ m ≤ 1000, 1 ≤ b ≤ 10000).

Output

Print the maximum number of bananas Okabe can get from the trees he cuts.

Examples

input

1 5

output

30

input

2 3

output

25

Note

The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas.

 给你一条经过1，3，4象限的直线，让我找到一个矩形价值最大的值，一个矩形的价值是这个矩形中所有的点X和Y的加和，这是个等差数列的和

直接暴力枚举好了

#include 
            
             #include 
             
              using namespace std;typedef long long LL;int main() {    int m,b;    scanf("%d%d",&m,&b);    LL ma=0;    for(int y=0; y<=b; y++) {        LL x=(b-y)*m;        LL t=x*(x+1)/2*(y+1)+y*(y+1)/2*(x+1);        ma=max(t,ma);    }    printf("%lld\n",ma);    return 0;}