#include <bits/stdc++.h>
using namespace std;
long long ar[2009][2009], pr[4009], ap[4009];
int main() {
  long long a, s, d, f, g, h, j, k, l, ans, flag, n, ans1, ans2, row1, row2,
      col1, col2;
  cin >> n;
  for (a = 1; a <= n; a++) {
    for (s = 1; s <= n; s++) {
      int temp;
      scanf("%d", &temp);
      ar[a][s] = temp;
      pr[a + s] += temp;
      ap[a + n - s - 1] += temp;
    }
  }
  ans1 = 0;
  ans2 = 0;
  for (a = 1; a <= n; a++) {
    for (s = 1; s <= n; s++) {
      if ((a + s) % 2) {
        if (pr[a + s] + ap[a + n - 1 - s] - ar[a][s] >= ans1) {
          ans1 = pr[a + s] + ap[a + n - 1 - s] - ar[a][s];
          row1 = a;
          col1 = s;
        }
      } else {
        if (pr[a + s] + ap[a + n - 1 - s] - ar[a][s] >= ans2) {
          ans2 = pr[a + s] + ap[a + n - 1 - s] - ar[a][s];
          row2 = a;
          col2 = s;
        }
      }
    }
  }
  cout << ans1 + ans2 << endl
       << row1 << " " << col1 << " " << row2 << " " << col2 << endl;
}

### 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 ar[2009][2009], pr[4009], ap[4009];
int main() {
  long long a, s, d, f, g, h, j, k, l, ans, flag, n, ans1, ans2, row1, row2,
      col1, col2;
  cin >> n;
  for (a = 1; a <= n; a++) {
    for (s = 1; s <= n; s++) {
      int temp;
      scanf("%d", &temp);
      ar[a][s] = temp;
      pr[a + s] += temp;
      ap[a + n - s - 1] += temp;
    }
  }
  ans1 = 0;
  ans2 = 0;
  for (a = 1; a <= n; a++) {
    for (s = 1; s <= n; s++) {
      if ((a + s) % 2) {
        if (pr[a + s] + ap[a + n - 1 - s] - ar[a][s] >= ans1) {
          ans1 = pr[a + s] + ap[a + n - 1 - s] - ar[a][s];
          row1 = a;
          col1 = s;
        }
      } else {
        if (pr[a + s] + ap[a + n - 1 - s] - ar[a][s] >= ans2) {
          ans2 = pr[a + s] + ap[a + n - 1 - s] - ar[a][s];
          row2 = a;
          col2 = s;
        }
      }
    }
  }
  cout << ans1 + ans2 << endl
       << row1 << " " << col1 << " " << row2 << " " << col2 << endl;
}

```