Problem

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.

The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15 

水，直接水过。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
int a[110][110];
int n,x;

int main()
{
    int sum,maxn;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&x);
                a[i][j]=a[i-1][j]+x;
            }
        }
        maxn=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i;j<=n;j++)
            {
                sum=0;
                for(int k=1;k<=n;k++)
                {
                    int t=a[j][k]-a[i-1][k];
                    sum+=t;
//                    cout<<"fuck "<<sum<<endl;
                    if(sum<0)
                        sum=0;
                    if(sum>maxn)
                        maxn=sum;
                }
            }
        }
        printf("%d\n",maxn);
    }
    return 0;
}