#include <bits/stdc++.h>
int a[2009][2009];
using namespace std;
long long d1[5000], d2[5009];
int main() {
  int n, i, j, k, x1, x2, y1, y2, i1, j1, k1;
  long long sum1 = -1, sum2 = -1, currsum = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      scanf("%d", &a[i][j]);
      d1[i + n - j] += (long long)a[i][j];
      d2[i + j] += (long long)a[i][j];
    }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      currsum = d1[i + n - j] + d2[i + j] - a[i][j];
      if ((i % 2 && j % 2) || (i % 2 == 0 && j % 2 == 0)) {
        if (currsum > sum1) {
          sum1 = currsum;
          x1 = i;
          y1 = j;
        }
      } else {
        if (currsum > sum2) {
          sum2 = currsum;
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << sum1 + sum2 << endl << x1 << " " << y1 << " " << x2 << " " << y2;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
int a[2009][2009];
using namespace std;
long long d1[5000], d2[5009];
int main() {
  int n, i, j, k, x1, x2, y1, y2, i1, j1, k1;
  long long sum1 = -1, sum2 = -1, currsum = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      scanf("%d", &a[i][j]);
      d1[i + n - j] += (long long)a[i][j];
      d2[i + j] += (long long)a[i][j];
    }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      currsum = d1[i + n - j] + d2[i + j] - a[i][j];
      if ((i % 2 && j % 2) || (i % 2 == 0 && j % 2 == 0)) {
        if (currsum > sum1) {
          sum1 = currsum;
          x1 = i;
          y1 = j;
        }
      } else {
        if (currsum > sum2) {
          sum2 = currsum;
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << sum1 + sum2 << endl << x1 << " " << y1 << " " << x2 << " " << y2;
  return 0;
}

```