#include <bits/stdc++.h>
using namespace std;
int n;
long long a[3000][3000] = {0};
long long b[3000][3000] = {0};
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0, input; j < n; j++) {
      scanf("%d", &input);
      a[i][j] = input;
    }
  }
  for (int i = 0; i < n; i++) {
    long long oni = 0;
    for (int j = 0; j < i + 1; j++) {
      oni += a[i - j][j];
    }
    for (int j = 0; j < i + 1; j++) {
      b[i - j][j] += oni;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    long long oni = 0;
    for (int j = 0; j < n - i; j++) {
      oni += a[i + j][n - j - 1];
    }
    for (int j = 0; j < n - i; j++) {
      b[i + j][n - j - 1] += oni;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    long long oni = 0;
    for (int j = 0; j < n - i; j++) {
      oni += a[i + j][j];
    }
    for (int j = 0; j < n - i; j++) {
      b[i + j][j] += oni;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    long long oni = 0;
    for (int j = 0; j <= i; j++) {
      oni += a[i - j][n - j - 1];
    }
    for (int j = 0; j <= i; j++) {
      b[i - j][n - j - 1] += oni;
    }
  }
  long long max1 = -1;
  int x1 = -1, y1 = -1;
  long long max2 = -1;
  int x2 = -1, y2 = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      b[i][j] -= a[i][j];
      if ((i + j) & 1) {
        if (max1 < b[i][j]) {
          max1 = b[i][j];
          x1 = i;
          y1 = j;
        }
      } else {
        if (max2 < b[i][j]) {
          max2 = b[i][j];
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << max1 + max2 << endl;
  cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1 << endl;
  return 0;
}

### 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;
int n;
long long a[3000][3000] = {0};
long long b[3000][3000] = {0};
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0, input; j < n; j++) {
      scanf("%d", &input);
      a[i][j] = input;
    }
  }
  for (int i = 0; i < n; i++) {
    long long oni = 0;
    for (int j = 0; j < i + 1; j++) {
      oni += a[i - j][j];
    }
    for (int j = 0; j < i + 1; j++) {
      b[i - j][j] += oni;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    long long oni = 0;
    for (int j = 0; j < n - i; j++) {
      oni += a[i + j][n - j - 1];
    }
    for (int j = 0; j < n - i; j++) {
      b[i + j][n - j - 1] += oni;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    long long oni = 0;
    for (int j = 0; j < n - i; j++) {
      oni += a[i + j][j];
    }
    for (int j = 0; j < n - i; j++) {
      b[i + j][j] += oni;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    long long oni = 0;
    for (int j = 0; j <= i; j++) {
      oni += a[i - j][n - j - 1];
    }
    for (int j = 0; j <= i; j++) {
      b[i - j][n - j - 1] += oni;
    }
  }
  long long max1 = -1;
  int x1 = -1, y1 = -1;
  long long max2 = -1;
  int x2 = -1, y2 = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      b[i][j] -= a[i][j];
      if ((i + j) & 1) {
        if (max1 < b[i][j]) {
          max1 = b[i][j];
          x1 = i;
          y1 = j;
        }
      } else {
        if (max2 < b[i][j]) {
          max2 = b[i][j];
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << max1 + max2 << endl;
  cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1 << endl;
  return 0;
}

```