#include <bits/stdc++.h>
using namespace std;
int main() {
  ios ::sync_with_stdio(false);
  int n, i, j, a, b;
  cin >> n;
  long long int arr[n + 1][n + 1], diagonal1[2 * n + 1], diagonal2[2 * n + 1];
  map<pair<int, int>, int> m1, m2;
  for (long long int i = 1; i < n + 1; i++)
    for (long long int j = 1; j < n + 1; j++) {
      cin >> arr[i][j];
    }
  for (long long int i = 0; i < 2 * n + 1; i++) diagonal1[i] = diagonal2[i] = 0;
  for (long long int i = 1; i < n + 1; i++) {
    for (long long int j = 1; j < n + 1; j++) {
      diagonal1[i + j] += arr[i][j];
      diagonal2[i - j + n] += arr[i][j];
    }
  }
  long long int oddmax = INT_MIN, evenmax = INT_MIN;
  int ex, ey, ox, oy;
  for (long long int i = 1; i < n + 1; i++) {
    for (long long int j = 1; j < n + 1; j++) {
      if ((i + j) % 2 == 1 &&
          oddmax < diagonal1[i + j] + diagonal2[i - j + n] - arr[i][j]) {
        oddmax = diagonal1[i + j] + diagonal2[i - j + n] - arr[i][j];
        ox = i;
        oy = j;
      } else if ((i + j) % 2 == 0 && evenmax < diagonal1[i + j] +
                                                   diagonal2[i - j + n] -
                                                   arr[i][j]) {
        evenmax = diagonal1[i + j] + diagonal2[i - j + n] - arr[i][j];
        ex = i;
        ey = j;
      }
    }
  }
  cout << oddmax + evenmax << "\n" << ox << " " << oy << " " << ex << " " << ey;
}

### 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);
  int n, i, j, a, b;
  cin >> n;
  long long int arr[n + 1][n + 1], diagonal1[2 * n + 1], diagonal2[2 * n + 1];
  map<pair<int, int>, int> m1, m2;
  for (long long int i = 1; i < n + 1; i++)
    for (long long int j = 1; j < n + 1; j++) {
      cin >> arr[i][j];
    }
  for (long long int i = 0; i < 2 * n + 1; i++) diagonal1[i] = diagonal2[i] = 0;
  for (long long int i = 1; i < n + 1; i++) {
    for (long long int j = 1; j < n + 1; j++) {
      diagonal1[i + j] += arr[i][j];
      diagonal2[i - j + n] += arr[i][j];
    }
  }
  long long int oddmax = INT_MIN, evenmax = INT_MIN;
  int ex, ey, ox, oy;
  for (long long int i = 1; i < n + 1; i++) {
    for (long long int j = 1; j < n + 1; j++) {
      if ((i + j) % 2 == 1 &&
          oddmax < diagonal1[i + j] + diagonal2[i - j + n] - arr[i][j]) {
        oddmax = diagonal1[i + j] + diagonal2[i - j + n] - arr[i][j];
        ox = i;
        oy = j;
      } else if ((i + j) % 2 == 0 && evenmax < diagonal1[i + j] +
                                                   diagonal2[i - j + n] -
                                                   arr[i][j]) {
        evenmax = diagonal1[i + j] + diagonal2[i - j + n] - arr[i][j];
        ex = i;
        ey = j;
      }
    }
  }
  cout << oddmax + evenmax << "\n" << ox << " " << oy << " " << ex << " " << ey;
}

```