#include <bits/stdc++.h>
using namespace std;
long long int sumr[2010][2010], suml[2010][2010], a[2010][2010];
int main() {
  ios::sync_with_stdio(false);
  long long int n, i, j, x1 = 1, x2 = 1, y1 = 1, y2 = 2;
  cin >> n;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      sumr[i][j] = sumr[i - 1][j - 1] + a[i][j];
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = n; j >= 1; j--) {
      suml[i][j] = suml[i - 1][j + 1] + a[i][j];
    }
  }
  long long int val1 = 0, val2 = 0, right = 0, left = 0;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (j >= i) {
        right = sumr[n - (j - i)][n];
      } else {
        right = sumr[n][n - (i - j)];
      }
      if ((n - j + 1) >= i) {
        left = suml[n - ((n - j + 1) - i)][1];
      } else {
        left = suml[n][n - (n - (i - (n - j + 1))) + 1];
      }
      if ((i + j) % 2 == 0) {
        if (right + left - a[i][j] > val1) {
          x1 = i;
          y1 = j;
          val1 = right + left - a[i][j];
        }
      } else {
        if (right + left - a[i][j] > val2) {
          x2 = i;
          y2 = j;
          val2 = right + left - a[i][j];
        }
      }
    }
  }
  cout << val1 + val2 << "\n";
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << "\n";
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long int sumr[2010][2010], suml[2010][2010], a[2010][2010];
int main() {
  ios::sync_with_stdio(false);
  long long int n, i, j, x1 = 1, x2 = 1, y1 = 1, y2 = 2;
  cin >> n;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      sumr[i][j] = sumr[i - 1][j - 1] + a[i][j];
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = n; j >= 1; j--) {
      suml[i][j] = suml[i - 1][j + 1] + a[i][j];
    }
  }
  long long int val1 = 0, val2 = 0, right = 0, left = 0;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (j >= i) {
        right = sumr[n - (j - i)][n];
      } else {
        right = sumr[n][n - (i - j)];
      }
      if ((n - j + 1) >= i) {
        left = suml[n - ((n - j + 1) - i)][1];
      } else {
        left = suml[n][n - (n - (i - (n - j + 1))) + 1];
      }
      if ((i + j) % 2 == 0) {
        if (right + left - a[i][j] > val1) {
          x1 = i;
          y1 = j;
          val1 = right + left - a[i][j];
        }
      } else {
        if (right + left - a[i][j] > val2) {
          x2 = i;
          y2 = j;
          val2 = right + left - a[i][j];
        }
      }
    }
  }
  cout << val1 + val2 << "\n";
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << "\n";
}

```