题目

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

3
4 0 2
3 5 7
8 1 6

output

9

input

4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1

output

1

input

4
1 1 1 1
1 1 0 1
1 1 2 1
1 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.

题解

在 0 的位置填入一个数,使每一行每一列以及两条最长的对角线的和相同
在输入的同时记录 0 的位置
找到任意一个没有 0 的行(列),计算总和,然后反推出 0 应该的取值
给每个 0 影响到的行加上这个值,判断一遍有没有不合法的即可

要注意

1
0

这组数据应该输出任意一个整数

计算时,由于 '0' 对总和没影响,可以在读入的同时直接记录下每行每列的和

判断总和只需要看第 0 行和第 1 行即可
选择 0 不在的那行就行
具体看代码

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

const int maxn = 505;
const int INF = 0x7FFFFFFF;

typedef __int64 LL;

LL Map[maxn][maxn];
LL sum1[maxn];
LL sum2[maxn];
LL sum3,sum4;

bool  Do() {
    int n;
    if(!(cin >> n))
        return false;

    int tx,ty;

    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= n;j++) {
            cin >> Map[i][j];
            if(Map[i][j] == 0) {
                tx = i;
                ty = j;
            }
        }
    LL ans = -1;
    bool can = true;

    if(n == 1) {
        ans = 1;
        can = true;
    } else {
        memset(sum1,0,sizeof(sum1));
        memset(sum2,0,sizeof(sum2));
        sum3 = sum4 = 0;

        for(int i = 1;i <= n;i++) {
            sum3 += Map[i][i];
            sum4 += Map[i][n - i + 1];
            for(int j = 1;j <= n;j++) {
                sum1[i] += Map[i][j];
                sum2[j] += Map[i][j];
            }
        }


        LL sum = 0;
        if(tx == 1)
            sum = sum1[2];
        else
            sum = sum1[1];

        ans = sum - sum1[tx];
        sum1[tx] += ans;
        sum2[ty] += ans;
        if(tx == ty)
            sum3 += ans;
        if(tx == n - ty + 1)
            sum4 += ans;

        for(int i = 1;i <= n;i++) {
            if(sum1[i] != sum || sum2[i] != sum) {
                can = false;
                break;
            }
        }
        if(sum3 != sum || sum4 != sum)
            can = false;
    }

    if(can && ans > 0)
        cout << ans << endl;
    else
        cout << "-1" << endl;

    return true;
}

int main() {
    while(Do());
    return 0;
}