#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, i, j, k, ans, white, black, x, x1, y1, x2, y2;
  cin >> n;
  vector<vector<long long>> a(n, vector<long long>(n));
  vector<long long> sx(n * 2 + 10, 0), sy(n * 2 + 10, 0);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
      sx[i + j] += a[i][j];
      sy[j - i + n] += a[i][j];
    }
  }
  white = 0;
  black = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      x = sx[i + j] + sy[j - i + n] - a[i][j];
      if ((i + j) % 2 == 0) {
        if (x >= white) {
          white = max(white, x);
          x1 = i;
          y1 = j;
        }
      } else {
        if (x >= black) {
          black = max(black, x);
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  ans = white + black;
  cout << ans << "\n";
  cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1 << "\n";
  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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, i, j, k, ans, white, black, x, x1, y1, x2, y2;
  cin >> n;
  vector<vector<long long>> a(n, vector<long long>(n));
  vector<long long> sx(n * 2 + 10, 0), sy(n * 2 + 10, 0);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
      sx[i + j] += a[i][j];
      sy[j - i + n] += a[i][j];
    }
  }
  white = 0;
  black = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      x = sx[i + j] + sy[j - i + n] - a[i][j];
      if ((i + j) % 2 == 0) {
        if (x >= white) {
          white = max(white, x);
          x1 = i;
          y1 = j;
        }
      } else {
        if (x >= black) {
          black = max(black, x);
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  ans = white + black;
  cout << ans << "\n";
  cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1 << "\n";
  return 0;
}

```