#include <bits/stdc++.h>
using namespace std;
const int NMAX = 2014;
long long int diag1[2 * NMAX], diag2[2 * NMAX];
long long int a[NMAX][NMAX];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  long long int x1, y1, x2, y2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
      diag1[i + j] += a[i][j];
      diag2[n + i - j] += a[i][j];
    }
  }
  long long int sol[2] = {-1, -1};
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (diag1[i + j] + diag2[i - j + n] - a[i][j] > sol[(i + j) & 1]) {
        sol[(i + j) & 1] = diag1[i + j] + diag2[i - j + n] - a[i][j];
        if ((i + j) % 2 == 0) {
          x1 = i;
          y1 = j;
        } else {
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << sol[0] + sol[1] << "\n";
  if (x1 < x2) {
    cout << x1 << " " << y1 << " " << x2 << " " << y2 << " ";
  } else {
    cout << x2 << " " << y2 << " " << x1 << " " << y1 << "\n";
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int NMAX = 2014;
long long int diag1[2 * NMAX], diag2[2 * NMAX];
long long int a[NMAX][NMAX];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  long long int x1, y1, x2, y2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
      diag1[i + j] += a[i][j];
      diag2[n + i - j] += a[i][j];
    }
  }
  long long int sol[2] = {-1, -1};
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (diag1[i + j] + diag2[i - j + n] - a[i][j] > sol[(i + j) & 1]) {
        sol[(i + j) & 1] = diag1[i + j] + diag2[i - j + n] - a[i][j];
        if ((i + j) % 2 == 0) {
          x1 = i;
          y1 = j;
        } else {
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << sol[0] + sol[1] << "\n";
  if (x1 < x2) {
    cout << x1 << " " << y1 << " " << x2 << " " << y2 << " ";
  } else {
    cout << x2 << " " << y2 << " " << x1 << " " << y1 << "\n";
  }
  return 0;
}

```