#include <bits/stdc++.h>
using namespace std;
inline int input() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int ret = 0;
  while (c >= '0' && c <= '9') {
    ret = 10 * ret + c - 48;
    c = getchar();
  }
  return ret;
}
long long int b[2001][2001];
int a[2001][2001];
int main() {
  int i, j, k, l, m, t, n;
  long long int count = 0, flag = 0;
  long long int ans = 0, x1, y1, x2, y2, maxi = INT_MIN, maxi1 = INT_MIN;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  i = 1;
  while (i <= n) {
    j = 1;
    k = i;
    count = 0;
    while (k <= n) {
      count += a[k][j];
      k++;
      j++;
    }
    k = i;
    j = 1;
    while (k <= n) {
      b[k][j] = count;
      k++;
      j++;
    }
    i += 2;
  }
  j = 3;
  while (j <= n) {
    i = 1;
    k = j;
    count = 0;
    while (k <= n) {
      count += a[i][k];
      i++;
      k++;
    }
    k = j;
    i = 1;
    while (k <= n) {
      b[i][k] = count;
      k++;
      i++;
    }
    j += 2;
  }
  j = 1;
  while (j <= n) {
    i = 1;
    k = j;
    count = 0;
    while (k >= 1) {
      count += a[i][k];
      k--;
      i++;
    }
    i = 1;
    k = j;
    while (k >= 1) {
      b[i][k] += (count - a[i][k]);
      if (b[i][k] > maxi) {
        maxi = b[i][k];
        x1 = i;
        y1 = k;
      }
      k--;
      i++;
    }
    j += 2;
  }
  if (j == n + 2)
    i = 3;
  else
    i = 2;
  while (i <= n) {
    k = i;
    j = n;
    count = 0;
    while (k <= n) {
      count += (a[k][j]);
      k++;
      j--;
    }
    k = i;
    j = n;
    while (k <= n) {
      b[k][j] += (count - a[k][j]);
      if (b[k][j] > maxi) {
        maxi = b[k][j];
        x1 = k;
        y1 = j;
      }
      k++;
      j--;
    }
    i += 2;
  }
  i = 2;
  while (i <= n) {
    j = 1;
    k = i;
    count = 0;
    while (k <= n) {
      count += a[k][j];
      k++;
      j++;
    }
    k = i;
    j = 1;
    while (k <= n) {
      b[k][j] = count;
      k++;
      j++;
    }
    i += 2;
  }
  j = 2;
  while (j <= n) {
    i = 1;
    k = j;
    count = 0;
    while (k <= n) {
      count += a[i][k];
      i++;
      k++;
    }
    k = j;
    i = 1;
    while (k <= n) {
      b[i][k] = count;
      k++;
      i++;
    }
    j += 2;
  }
  j = 2;
  while (j <= n) {
    i = 1;
    k = j;
    count = 0;
    while (k >= 1) {
      count += a[i][k];
      k--;
      i++;
    }
    i = 1;
    k = j;
    while (k >= 1) {
      b[i][k] += (count - a[i][k]);
      if (b[i][k] > maxi1) {
        maxi1 = b[i][k];
        x2 = i;
        y2 = k;
      }
      k--;
      i++;
    }
    j += 2;
  }
  if (j == n + 2)
    i = 3;
  else
    i = 2;
  while (i <= n) {
    k = i;
    j = n;
    count = 0;
    while (k <= n) {
      count += (a[k][j]);
      k++;
      j--;
    }
    k = i;
    j = n;
    while (k <= n) {
      b[k][j] += (count - a[k][j]);
      if (b[k][j] > maxi1) {
        maxi1 = b[k][j];
        x2 = k;
        y2 = j;
      }
      k++;
      j--;
    }
    i += 2;
  }
  cout << maxi + maxi1 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

Gargari is jealous that his friend Caisa won the game from the previous problem. He wants to prove that he is a genius.

He has a n × n chessboard. Each cell of the chessboard has a number written on it. Gargari wants to place two bishops on the chessboard in such a way that there is no cell that is attacked by both of them. Consider a cell with number x written on it, if this cell is attacked by one of the bishops Gargari will get x dollars for it. Tell Gargari, how to place bishops on the chessboard to get maximum amount of money.

We assume a cell is attacked by a bishop, if the cell is located on the same diagonal with the bishop (the cell, where the bishop is, also considered attacked by it).

Input

The first line contains a single integer n (2 ≤ n ≤ 2000). Each of the next n lines contains n integers aij (0 ≤ aij ≤ 109) — description of the chessboard.

Output

On the first line print the maximal number of dollars Gargari will get. On the next line print four integers: x1, y1, x2, y2 (1 ≤ x1, y1, x2, y2 ≤ n), where xi is the number of the row where the i-th bishop should be placed, yi is the number of the column where the i-th bishop should be placed. Consider rows are numbered from 1 to n from top to bottom, and columns are numbered from 1 to n from left to right.

If there are several optimal solutions, you can print any of them.

Examples

Input

4
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1


Output

12
2 2 3 2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline int input() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int ret = 0;
  while (c >= '0' && c <= '9') {
    ret = 10 * ret + c - 48;
    c = getchar();
  }
  return ret;
}
long long int b[2001][2001];
int a[2001][2001];
int main() {
  int i, j, k, l, m, t, n;
  long long int count = 0, flag = 0;
  long long int ans = 0, x1, y1, x2, y2, maxi = INT_MIN, maxi1 = INT_MIN;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  i = 1;
  while (i <= n) {
    j = 1;
    k = i;
    count = 0;
    while (k <= n) {
      count += a[k][j];
      k++;
      j++;
    }
    k = i;
    j = 1;
    while (k <= n) {
      b[k][j] = count;
      k++;
      j++;
    }
    i += 2;
  }
  j = 3;
  while (j <= n) {
    i = 1;
    k = j;
    count = 0;
    while (k <= n) {
      count += a[i][k];
      i++;
      k++;
    }
    k = j;
    i = 1;
    while (k <= n) {
      b[i][k] = count;
      k++;
      i++;
    }
    j += 2;
  }
  j = 1;
  while (j <= n) {
    i = 1;
    k = j;
    count = 0;
    while (k >= 1) {
      count += a[i][k];
      k--;
      i++;
    }
    i = 1;
    k = j;
    while (k >= 1) {
      b[i][k] += (count - a[i][k]);
      if (b[i][k] > maxi) {
        maxi = b[i][k];
        x1 = i;
        y1 = k;
      }
      k--;
      i++;
    }
    j += 2;
  }
  if (j == n + 2)
    i = 3;
  else
    i = 2;
  while (i <= n) {
    k = i;
    j = n;
    count = 0;
    while (k <= n) {
      count += (a[k][j]);
      k++;
      j--;
    }
    k = i;
    j = n;
    while (k <= n) {
      b[k][j] += (count - a[k][j]);
      if (b[k][j] > maxi) {
        maxi = b[k][j];
        x1 = k;
        y1 = j;
      }
      k++;
      j--;
    }
    i += 2;
  }
  i = 2;
  while (i <= n) {
    j = 1;
    k = i;
    count = 0;
    while (k <= n) {
      count += a[k][j];
      k++;
      j++;
    }
    k = i;
    j = 1;
    while (k <= n) {
      b[k][j] = count;
      k++;
      j++;
    }
    i += 2;
  }
  j = 2;
  while (j <= n) {
    i = 1;
    k = j;
    count = 0;
    while (k <= n) {
      count += a[i][k];
      i++;
      k++;
    }
    k = j;
    i = 1;
    while (k <= n) {
      b[i][k] = count;
      k++;
      i++;
    }
    j += 2;
  }
  j = 2;
  while (j <= n) {
    i = 1;
    k = j;
    count = 0;
    while (k >= 1) {
      count += a[i][k];
      k--;
      i++;
    }
    i = 1;
    k = j;
    while (k >= 1) {
      b[i][k] += (count - a[i][k]);
      if (b[i][k] > maxi1) {
        maxi1 = b[i][k];
        x2 = i;
        y2 = k;
      }
      k--;
      i++;
    }
    j += 2;
  }
  if (j == n + 2)
    i = 3;
  else
    i = 2;
  while (i <= n) {
    k = i;
    j = n;
    count = 0;
    while (k <= n) {
      count += (a[k][j]);
      k++;
      j--;
    }
    k = i;
    j = n;
    while (k <= n) {
      b[k][j] += (count - a[k][j]);
      if (b[k][j] > maxi1) {
        maxi1 = b[k][j];
        x2 = k;
        y2 = j;
      }
      k++;
      j--;
    }
    i += 2;
  }
  cout << maxi + maxi1 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
  return 0;
}

```