Chris and Magic Square CodeForces - 711B

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.

Example：

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.

//可以说是考逻辑的题吧,特别注意的是要特判n=1的时候的情况
//先把数独的每行每列的和存储在两个数组中,然后找出最大和最小值的差ans,
//如果最大值减去最小值小于0的话,则证明不存在一个数可以满足条件,则输出-1,
//如果ans大于0的话,就把ans补在那个为0的地方,注意这是要用两个空数组把每一行和每一列的和
//存起来,可以把之前的b,c两个数组清0存储,然后再把两个对角线加起来,分别判断两个对角线的值是否
//和之前的最大值相等,b[i]和c[i]是否和最大值相等
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const int INF=1e9+7;
const int maxn=510;
typedef long long ll;

int  n;
ll b[maxn],c[maxn];
ll a[maxn][maxn];
ll x,y,ans,minn,maxx,sum1,sum2;

int main()
{
    scanf("%d",&n);
    ll i,j;
    ans=sum1=sum2=0;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    for(i=0; i<n; i++)
    {
        for(j=0; j<n; j++)
        {
            scanf("%I64d",&a[i][j]);
            b[i]+=a[i][j];
            c[j]+=a[i][j];
            if(a[i][j]==0)
            {
                x=i;
                y=j;
            }
        }
    }
    minn=maxx=0;
    for(i=0; i<n; i++)
    {
        if(i!=x) maxx=b[i];
        else minn=b[i];
    }
    ans=maxx-minn;
    if(n==1)
    {
        printf("1\n");
        return 0;
    }
    if(ans<=0)
    {
        printf("-1\n");
        return 0;
    }
    a[x][y]=ans;
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    for(i=0; i<n; i++)
    {
        for(j=0; j<n; j++)
        {
            if(i==j) sum1+=a[i][j];
            if(i==n-1-j)sum2+=a[i][j];
            b[i]+=a[i][j];
            c[j]+=a[i][j];
        }
    }

    if(sum1!=maxx||sum2!=maxx)
    {
        printf("-1\n");
        return 0;
    }
    for(i=0; i<n; i++)
    {
        if(b[i]!=maxx||c[i]!=maxx)
        {
            printf("-1\n");
            return 0;
        }
    }
    printf("%I64d\n",ans);
    return 0;
}