B. Working out

Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.

Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].

There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.

If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.

Input

The first line of the input contains two integers n and m (3 ≤ n, m ≤ 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 ≤ a[i][j] ≤ 105).

Output

The output contains a single number — the maximum total gain possible.

Examples

input

Copy

3 3
100 100 100
100 1 100
100 100 100

output

Copy

800

Note

Iahub will choose exercises a[1][1] → a[1][2] → a[2][2] → a[3][2] → a[3][3]. Iahubina will choose exercises a[3][1] → a[2][1] → a[2][2] → a[2][3] → a[1][3].

题意:

有个n,m的矩阵，一个从（1,1）出发到(n,m),一个从(n,1)出发到(n,m),他们方向只要两个分别不同，一个向右，向下，一个向上，向右。然后他们只能相遇一次(这个很关键)，相遇的格子里的值不计算，求两人经过格子的最大值。

思路：

通过分析可知，一个人从上进入，那他必定从下面出，一个从左边进，必定从右边出。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=1e3+100;
LL a[maxn][maxn];
LL dp1[maxn][maxn],dp2[maxn][maxn],dp3[maxn][maxn],dp4[maxn][maxn];
void init(int n,int m)
{
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            dp1[i][j]=max(dp1[i-1][j]+a[i][j],dp1[i][j-1]+a[i][j]);
        }
    }
    for(int i=n;i>=1;i--)
    {
        for(int j=m;j>=1;j--)
        {
            dp2[i][j]=max(dp2[i+1][j]+a[i][j],dp2[i][j+1]+a[i][j]);
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=m;j>=1;j--)
        {
            dp3[i][j]=max(dp3[i-1][j]+a[i][j],dp3[i][j+1]+a[i][j]);
        }
    }
    for(int i=n;i>=1;i--)
    {
        for(int j=1;j<=m;j++)
        {
            dp4[i][j]=max(dp4[i+1][j]+a[i][j],dp4[i][j-1]+a[i][j]);
        }
    }
    return ;
}
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>a[i][j];
        }
    }
    init(n,m);
    LL ans=-1;
    for(int i=2;i<n;i++)
    {
        for(int j=2;j<m;j++)
        {
            ans=max(ans,dp1[i-1][j]+dp2[i+1][j]+dp3[i][j+1]+dp4[i][j-1]);
            ans=max(ans,dp1[i][j-1]+dp2[i][j+1]+dp3[i-1][j]+dp4[i+1][j]);
        }
    }
    cout<<ans<<endl;
    return 0;
}