#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, i, j, x1 = 1, x2 = 1, y1 = 1, y2 = 2, max1 = 0, max2 = 0;
  cin >> n;
  long long int a[n][n], x[2 * n - 1], y[2 * n - 1];
  for (i = 0; i < 2 * n - 1; i++) {
    x[i] = 0;
    y[i] = 0;
  }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
      x[i + j] += a[i][j];
      y[n - 1 - i + j] += a[i][j];
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      if ((i + j) % 2) {
        if (max1 < x[i + j] + y[n - i + j - 1] - a[i][j]) {
          max1 = x[i + j] + y[n - i + j - 1] - a[i][j];
          x1 = i + 1;
          y1 = j + 1;
        }
      } else {
        if (max2 < x[i + j] + y[n - i + j - 1] - a[i][j]) {
          max2 = x[i + j] + y[n - i + j - 1] - a[i][j];
          x2 = i + 1;
          y2 = j + 1;
        }
      }
    }
  cout << max1 + max2 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
}

### Prompt

Your challenge is to write a CPP solution to 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(NULL);
  long long int n, i, j, x1 = 1, x2 = 1, y1 = 1, y2 = 2, max1 = 0, max2 = 0;
  cin >> n;
  long long int a[n][n], x[2 * n - 1], y[2 * n - 1];
  for (i = 0; i < 2 * n - 1; i++) {
    x[i] = 0;
    y[i] = 0;
  }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
      x[i + j] += a[i][j];
      y[n - 1 - i + j] += a[i][j];
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      if ((i + j) % 2) {
        if (max1 < x[i + j] + y[n - i + j - 1] - a[i][j]) {
          max1 = x[i + j] + y[n - i + j - 1] - a[i][j];
          x1 = i + 1;
          y1 = j + 1;
        }
      } else {
        if (max2 < x[i + j] + y[n - i + j - 1] - a[i][j]) {
          max2 = x[i + j] + y[n - i + j - 1] - a[i][j];
          x2 = i + 1;
          y2 = j + 1;
        }
      }
    }
  cout << max1 + max2 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
}

```