#include <bits/stdc++.h>
using namespace std;
using vi = vector<long long int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vc = vector<char>;
using vs = vector<string>;
using vld = vector<long double>;
using pii = pair<long long int, long long int>;
using psi = pair<string, long long int>;
using pci = pair<char, long long int>;
using vpii = vector<pii>;
long long int mod = 1e9 + 7;
long long int const maxn = 1e5 + 5;
long long int const inf = 1e18;
long long int add(long long int a, long long int b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long int mul(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int powm(long long int x, long long int n, long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long int modinverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
bool prime(long long int x) {
  if (x <= 1) return false;
  for (int i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return false;
  return true;
}
long long int divisor(long long int x) {
  long long int cnt = 0;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (i != x / i)
        cnt += 2;
      else
        cnt += 1;
    }
  }
  return cnt;
}
vector<long long int> sieve(long long int n) {
  bool prim[n + 1];
  memset(prim, true, sizeof(prim));
  for (long long int p = 2; p * p <= n; p++) {
    if (prim[p] == true) {
      for (int i = p * p; i <= n; i += p) prim[i] = false;
    }
  }
  vector<long long int> v;
  for (int i = 2; i <= n; i++)
    if (prim[i]) v.push_back(i);
  return v;
}
long long int const N = 1005;
long long int dp1[N][N];
long long int dp2[N][N];
long long int dp3[N][N];
long long int dp4[N][N];
long long int a[N][N];
void solve() {
  long long int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp1[i][j] = a[i][j] + max(dp1[i - 1][j], dp1[i][j - 1]);
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      dp2[i][j] = a[i][j] + max(dp2[i + 1][j], dp2[i][j + 1]);
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = 1; j <= m; j++) {
      dp3[i][j] = a[i][j] + max(dp3[i + 1][j], dp3[i][j - 1]);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = m; j >= 1; j--) {
      dp4[i][j] = a[i][j] + max(dp4[i - 1][j], dp4[i][j + 1]);
    }
  }
  long long int ans = 0;
  for (int i = 2; i < n; i++) {
    for (int j = 2; j < m; j++) {
      ans = max(ans,
                dp1[i][j - 1] + dp2[i][j + 1] + dp3[i + 1][j] + dp4[i - 1][j]);
      ans = max(ans,
                dp1[i - 1][j] + dp2[i + 1][j] + dp3[i][j - 1] + dp4[i][j + 1]);
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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

3 3
100 100 100
100 1 100
100 100 100


Output

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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using vi = vector<long long int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vc = vector<char>;
using vs = vector<string>;
using vld = vector<long double>;
using pii = pair<long long int, long long int>;
using psi = pair<string, long long int>;
using pci = pair<char, long long int>;
using vpii = vector<pii>;
long long int mod = 1e9 + 7;
long long int const maxn = 1e5 + 5;
long long int const inf = 1e18;
long long int add(long long int a, long long int b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long int mul(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int powm(long long int x, long long int n, long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long int modinverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
bool prime(long long int x) {
  if (x <= 1) return false;
  for (int i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return false;
  return true;
}
long long int divisor(long long int x) {
  long long int cnt = 0;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (i != x / i)
        cnt += 2;
      else
        cnt += 1;
    }
  }
  return cnt;
}
vector<long long int> sieve(long long int n) {
  bool prim[n + 1];
  memset(prim, true, sizeof(prim));
  for (long long int p = 2; p * p <= n; p++) {
    if (prim[p] == true) {
      for (int i = p * p; i <= n; i += p) prim[i] = false;
    }
  }
  vector<long long int> v;
  for (int i = 2; i <= n; i++)
    if (prim[i]) v.push_back(i);
  return v;
}
long long int const N = 1005;
long long int dp1[N][N];
long long int dp2[N][N];
long long int dp3[N][N];
long long int dp4[N][N];
long long int a[N][N];
void solve() {
  long long int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp1[i][j] = a[i][j] + max(dp1[i - 1][j], dp1[i][j - 1]);
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      dp2[i][j] = a[i][j] + max(dp2[i + 1][j], dp2[i][j + 1]);
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = 1; j <= m; j++) {
      dp3[i][j] = a[i][j] + max(dp3[i + 1][j], dp3[i][j - 1]);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = m; j >= 1; j--) {
      dp4[i][j] = a[i][j] + max(dp4[i - 1][j], dp4[i][j + 1]);
    }
  }
  long long int ans = 0;
  for (int i = 2; i < n; i++) {
    for (int j = 2; j < m; j++) {
      ans = max(ans,
                dp1[i][j - 1] + dp2[i][j + 1] + dp3[i + 1][j] + dp4[i - 1][j]);
      ans = max(ans,
                dp1[i - 1][j] + dp2[i + 1][j] + dp3[i][j - 1] + dp4[i][j + 1]);
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```