#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long a[n + 2][n + 2], l[4004] = {0}, r[4004] = {0}, ans1 = 0, ans2 = 0,
                             ans1i, ans1j, ans2i, ans2j;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
      l[i + j] += a[i][j];
      r[i - j + n] += a[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      if ((i + j) % 2 == 1) {
        if (ans1 <= l[i + j] + r[i - j + n] - a[i][j]) {
          ans1 = l[i + j] + r[i - j + n] - a[i][j];
          ans1i = i;
          ans1j = j;
        }
      } else {
        if (ans2 <= l[i + j] + r[i - j + n] - a[i][j]) {
          ans2 = l[i + j] + r[i - j + n] - a[i][j];
          ans2i = i;
          ans2j = j;
        }
      }
  }
  cout << ans1 + ans2 << endl;
  cout << ans1i << " " << ans1j << " " << ans2i << " " << ans2j;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long a[n + 2][n + 2], l[4004] = {0}, r[4004] = {0}, ans1 = 0, ans2 = 0,
                             ans1i, ans1j, ans2i, ans2j;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
      l[i + j] += a[i][j];
      r[i - j + n] += a[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      if ((i + j) % 2 == 1) {
        if (ans1 <= l[i + j] + r[i - j + n] - a[i][j]) {
          ans1 = l[i + j] + r[i - j + n] - a[i][j];
          ans1i = i;
          ans1j = j;
        }
      } else {
        if (ans2 <= l[i + j] + r[i - j + n] - a[i][j]) {
          ans2 = l[i + j] + r[i - j + n] - a[i][j];
          ans2i = i;
          ans2j = j;
        }
      }
  }
  cout << ans1 + ans2 << endl;
  cout << ans1i << " " << ans1j << " " << ans2i << " " << ans2j;
  return 0;
}

```