Codeforces 711B- Chris and Magic Square

B. Chris and Magic Square

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n × n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.

Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (), each column of the grid (), and the two long diagonals of the grid (the main diagonal —  and the secondary diagonal — ) are equal.

Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the number of rows and columns of the magic grid.

n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 ≤ ai, j ≤ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.

It is guaranteed that there is exactly one pair of integers i, j (1 ≤ i, j ≤ n) such that ai, j = 0.

Output

Output a single integer, the positive integer x (1 ≤ x ≤ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output  - 1 instead.

If there are multiple solutions, you may print any of them.

Examples

input

34 0 23 5 78 1 6

output

9

input

41 1 1 11 1 0 11 1 1 11 1 1 1

output

1

input

41 1 1 11 1 0 11 1 2 11 1 1 1

output

-1

Note

In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,

The sum of numbers in each row is:

4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.

The sum of numbers in each column is:

4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.

The sum of numbers in the two diagonals is:

4 + 5 + 6 = 2 + 5 + 8 = 15.

In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.

题意：给定一个二位数组，其中有一个是0，找出一个整数代替0使得数组的每一行，每一列以及对角线的和相等。如果该数字不存在输出-1。

解：在输入过程中记录0出现的位置x[ii][jj]以及出现的数量q，如果q>1或者n==1直接输出-1，反之求出每一行的和flagi以及每一列的和flagj，如果求得结果不同则不存在该整数，直接输出-1；如果相等，减去0存在的行的和a[0]或者列的和b[0]求得该整数count，将原先的0替换为count，求出对角线的和z和w，如果对角线的和跟行的和相等则输出count否则输出-1.

#include<stdio.h>#include<algorithm>#include<string.h>#include<iostream>#include<math.h>using namespace std;long long x[505][505];long long fin(long long a[],int n){long long flag=1;long long x=a[1];for(int i=1;i<n;i++){if(x!=a[i]){flag=0;}else {flag=x;}}return flag;}int main(){int n;int ii,jj;int q=0;ii=jj=0;long long a[505];long long b[505];long long ai=0;long long bj=0;long long count=0;memset(a,0,sizeof(a));memset(b,0,sizeof(b));scanf("%d",&n);for(int i=0;i<n;i++){for(int j=0;j<n;j++){scanf("%I64d",&x[i][j]);if(x[i][j]==0){ii=i;jj=j;q++;}}}for(int i=0;i<n;i++){for(int j=0;j<n;j++){a[i]+=x[i][j];}}for(int i=0;i<n;i++){for(int j=0;j<n;j++){b[i]+=x[j][i];}}sort(a,a+n);sort(b,b+n);long long flagi=fin(a,n);long long flagj=fin(b,n);if(q>1){printf("-1\n");}else if(n==1){printf("1\n");}else if(flagi!=0 && flagj!=0 && flagi==flagj && a[0]==b[0]){count=flagi-a[0];x[ii][jj]=count;int z=0;while(z<n){ai+=x[z][z];z++;}int w=n-1;while(w>=0){bj+=x[n-1-w][w];w--;}if(ai==bj && ai==flagi){if(count>0){printf("%I64d\n",count);}else{printf("-1\n");}}else {printf("-1\n");}}else {printf("-1\n");}return 0;}