#include <bits/stdc++.h>
using namespace std;
long long M[2001][2001], diaMain[2001][2001], dia2[2001][2001];
long long MAX(long long a, long long b) {
  if (a >= b) return a;
  return b;
}
int main() {
  long long anso = 0, anse = 0, a1 = 0, a2 = 0, x1 = 0, x2 = 1, y1 = 0, y2 = 0;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      scanf("%I64d", &M[i][j]);
      diaMain[i - min(i, j)][j - min(i, j)] += M[i][j];
      dia2[i - min(i, n - 1 - j)][n - 1 - j - min(i, n - 1 - j)] += M[i][j];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i + j) % 2 == 1) {
        anso =
            MAX(anso,
                diaMain[i - min(i, j)][j - min(i, j)] +
                    dia2[i - min(i, n - 1 - j)][n - 1 - j - min(i, n - 1 - j)] -
                    M[i][j]);
        if (anso > a1) {
          a1 = anso;
          x1 = i, y1 = j;
        }
      } else {
        anse =
            MAX(anse,
                diaMain[i - min(i, j)][j - min(i, j)] +
                    dia2[i - min(i, n - 1 - j)][n - 1 - j - min(i, n - 1 - j)] -
                    M[i][j]);
        if (anse > a2) {
          a2 = anse;
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  printf("%I64d\n", anso + anse);
  cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long M[2001][2001], diaMain[2001][2001], dia2[2001][2001];
long long MAX(long long a, long long b) {
  if (a >= b) return a;
  return b;
}
int main() {
  long long anso = 0, anse = 0, a1 = 0, a2 = 0, x1 = 0, x2 = 1, y1 = 0, y2 = 0;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      scanf("%I64d", &M[i][j]);
      diaMain[i - min(i, j)][j - min(i, j)] += M[i][j];
      dia2[i - min(i, n - 1 - j)][n - 1 - j - min(i, n - 1 - j)] += M[i][j];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i + j) % 2 == 1) {
        anso =
            MAX(anso,
                diaMain[i - min(i, j)][j - min(i, j)] +
                    dia2[i - min(i, n - 1 - j)][n - 1 - j - min(i, n - 1 - j)] -
                    M[i][j]);
        if (anso > a1) {
          a1 = anso;
          x1 = i, y1 = j;
        }
      } else {
        anse =
            MAX(anse,
                diaMain[i - min(i, j)][j - min(i, j)] +
                    dia2[i - min(i, n - 1 - j)][n - 1 - j - min(i, n - 1 - j)] -
                    M[i][j]);
        if (anse > a2) {
          a2 = anse;
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  printf("%I64d\n", anso + anse);
  cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1;
  return 0;
}

```