#include <bits/stdc++.h>
using namespace std;
long long d[4000], e[4000];
int a[2000][2000];
int main() {
  int n, i, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      scanf("%d", &a[i][j]), d[i + j] += a[i][j], e[i - j + n] += a[i][j];
  long long vv, y = -1, z = -1;
  int p, q, r, s;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      vv = d[i + j] + e[i - j + n] - a[i][j];
      if ((i + j) & 1) {
        if (vv > y) y = vv, p = i, q = j;
      } else {
        if (vv > z) z = vv, r = i, s = j;
      }
    }
  cout << y + z << endl
       << p + 1 << " " << q + 1 << " " << r + 1 << " " << s + 1 << endl;
}

### Prompt

Please formulate 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;
long long d[4000], e[4000];
int a[2000][2000];
int main() {
  int n, i, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      scanf("%d", &a[i][j]), d[i + j] += a[i][j], e[i - j + n] += a[i][j];
  long long vv, y = -1, z = -1;
  int p, q, r, s;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      vv = d[i + j] + e[i - j + n] - a[i][j];
      if ((i + j) & 1) {
        if (vv > y) y = vv, p = i, q = j;
      } else {
        if (vv > z) z = vv, r = i, s = j;
      }
    }
  cout << y + z << endl
       << p + 1 << " " << q + 1 << " " << r + 1 << " " << s + 1 << endl;
}

```