Okabe and Future Gadget Laboratory

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 tso 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.

Example

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".

#include
int main()
{
int n,i,j,a[55][55],s,t,x,y,flag;
while(scanf("%d",&n)!=EOF)
{
s=t=0;
for(i=0;ifor(j=0;jscanf("%d",&a[i][j]);
for(i=0;ifor(j=0;jif(a[i][j]!=1)
s++;
for(i=0;ifor(j=0;j{
flag=1;
for(x=0;x{
if(x!=j)
for(y=0;yif(y!=i)
if((a[y][j]+a[i][x])==a[i][j])
{
flag=0;
t++;
}
}
}
if(s==t)
printf("YES\n");
else
printf("NO\n");
}
return 0;
 }