Round #420 A. Okabe and Future Gadget Laboratory(Div.2)

Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an n by n square 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 a_x, y ≠ 1, there should exist two indices s and t so that a_x, y = a_x, s + a_t, y, where a_i, 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 a_i, j (1 ≤ a_i, 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".

想明白后才发现，答案原来题目中已有，笑哭呀，还是要好好听老师说的话，好好读题，答案就在题中。

题意：有一个n×n的格子，格子里面的数如果大于1，那么它就必须等于它所在列中的一个数加它所在行的一个数之和，

不满足就是No，

#include <iostream>
#include <stdio.h>
using namespace std;
int a[55][55];
int main(){
    int n,i,j;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
      for(j=1;j<=n;j++)
        cin>>a[i][j];//输入方格
    
    for(i=1;i<=n;i++)
      for(j=1;j<=n;j++)    
          if(a[i][j]!=1){ //大于1
          bool flag =false ;//定义一个标志
          for(int s=1;s<=n&&!flag;s++)   //就样例一（3,1）举例
              for(int t=1;t<=n&&!flag;t++)
                  flag=(a[i][j]==a[i][s]+a[t][j]); //（3，1）==（3,1）+（1,1），flag=false，t++,
                                                   //（3，1）==（3,1）+（2,1）……都不满足，返回上个循环，以此类推

          if(!flag) {
              printf("No
");return 0;
            }
          } 
    printf("Yes
");
    return 0; 
}