#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#include<bits/stdc++.h>
#define rc(x) return cout<<x<<endl,0
#define pb push_back
#define in insert
#define er erase
#define fd find
#define fr first
#define sc second
typedef long long ll;
typedef long double ld;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll llinf=(1LL<<62);
const int inf=(1<<30);
const int nmax=1e5+50;
const int mod=1e9+7;
using namespace std;
int n,m,i,v,u,x,y,t,p,j;
ll a[205][205],f[405][205][205];
int main()
{
	//freopen("sol.in","r",stdin);
	//freopen("sol.out","w",stdout);
	//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
	cin>>n>>m;
	for(i=1;i<=n;i++)for(j=1;j<=m;j++)cin>>a[i][j];
	for(i=1;i<=n+m-1;i++)
	{
		for(x=1;x<=n;x++)
		{
			for(y=1;y<=n;y++)
			{
				v=i-x+1,u=i-y+1;
				if(v<=0 || u<=0)continue;
				//cout<<x<<" "<<v<<" "<<y<<" "<<u<<endl;
				for(t=-1;t<1;t++)
				{
					for(p=-1;p<1;p++)
					{
						f[i][x][y]=max(f[i][x][y],f[i-1][x+t][y+p]);
					}
				}
				f[i][x][y]+=a[x][v]+a[y][u];
				if(x==y && v==u)f[i][x][y]-=a[x][v];
			}
		}
	}
	cout<<f[n+m-1][n][n]<<endl;
    return 0;
}
### Prompt

Create a solution in Cpp for the following problem:

Input

The input is given from standard input in the following format.


> $H \ W$ $a_{1, 1} \ a_{1, 2} \ \cdots \ a_{1, W}$ $a_{2, 1} \ a_{2, 2} \ \cdots \ a_{2, W}$ $\vdots \ \ \ \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \ \vdots$ $a_{H, 1} \ a_{H, 2} \ \cdots \ a_{H, W}$

Output

* Print the maximum number of souvenirs they can get.



Constraints

* $1 \le H, W \le 200$
* $0 \le a_{i, j} \le 10^5$



Subtasks

Subtask 1 [ 50 points ]


* The testcase in the subtask satisfies $1 \le H \le 2$.

Subtask 2 [ 80 points ]


* The testcase in the subtask satisfies $1 \le H \le 3$.

Subtask 3 [ 120 points ]


* The testcase in the subtask satisfies $1 \le H, W \le 7$.

Subtask 4 [ 150 points ]


* The testcase in the subtask satisfies $1 \le H, W \le 30$.

Subtask 5 [ 200 points ]


* There are no additional constraints.

Output

* Print the maximum number of souvenirs they can get.



Constraints

* $1 \le H, W \le 200$
* $0 \le a_{i, j} \le 10^5$



Subtasks

Subtask 1 [ 50 points ]


* The testcase in the subtask satisfies $1 \le H \le 2$.

Subtask 2 [ 80 points ]


* The testcase in the subtask satisfies $1 \le H \le 3$.

Subtask 3 [ 120 points ]


* The testcase in the subtask satisfies $1 \le H, W \le 7$.

Subtask 4 [ 150 points ]


* The testcase in the subtask satisfies $1 \le H, W \le 30$.

Subtask 5 [ 200 points ]


* There are no additional constraints.

Input

The input is given from standard input in the following format.


> $H \ W$ $a_{1, 1} \ a_{1, 2} \ \cdots \ a_{1, W}$ $a_{2, 1} \ a_{2, 2} \ \cdots \ a_{2, W}$ $\vdots \ \ \ \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \ \vdots$ $a_{H, 1} \ a_{H, 2} \ \cdots \ a_{H, W}$

Examples

Input

3 3
1 0 5
2 2 3
4 2 4


Output

21


Input

6 6
1 2 3 4 5 6
8 6 9 1 2 0
3 1 4 1 5 9
2 6 5 3 5 8
1 4 1 4 2 1
2 7 1 8 2 8


Output

97

### Response

```cpp
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#include<bits/stdc++.h>
#define rc(x) return cout<<x<<endl,0
#define pb push_back
#define in insert
#define er erase
#define fd find
#define fr first
#define sc second
typedef long long ll;
typedef long double ld;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll llinf=(1LL<<62);
const int inf=(1<<30);
const int nmax=1e5+50;
const int mod=1e9+7;
using namespace std;
int n,m,i,v,u,x,y,t,p,j;
ll a[205][205],f[405][205][205];
int main()
{
	//freopen("sol.in","r",stdin);
	//freopen("sol.out","w",stdout);
	//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
	cin>>n>>m;
	for(i=1;i<=n;i++)for(j=1;j<=m;j++)cin>>a[i][j];
	for(i=1;i<=n+m-1;i++)
	{
		for(x=1;x<=n;x++)
		{
			for(y=1;y<=n;y++)
			{
				v=i-x+1,u=i-y+1;
				if(v<=0 || u<=0)continue;
				//cout<<x<<" "<<v<<" "<<y<<" "<<u<<endl;
				for(t=-1;t<1;t++)
				{
					for(p=-1;p<1;p++)
					{
						f[i][x][y]=max(f[i][x][y],f[i-1][x+t][y+p]);
					}
				}
				f[i][x][y]+=a[x][v]+a[y][u];
				if(x==y && v==u)f[i][x][y]-=a[x][v];
			}
		}
	}
	cout<<f[n+m-1][n][n]<<endl;
    return 0;
}
```