#include <bits/stdc++.h>
using namespace std;
long long n, p, q, r, s;
long long b[2002][2002];
long long le[5000];
long long ri[5000];
int main() {
  scanf(" %I64d", &n);
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      scanf(" %I64d", &b[i][j]);
      le[i + j] += b[i][j];
      ri[n + i - j] += b[i][j];
    }
  }
  long long odd = -1, even = -1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      long long ans = le[i + j] + ri[n + i - j] - b[i][j];
      if ((i + j) & 1) {
        if (ans > odd) {
          odd = ans;
          p = i;
          q = j;
        }
      } else {
        if (ans > even) {
          even = ans;
          r = i;
          s = j;
        }
      }
    }
  }
  cout << even + odd << "\n" << p << " " << q << " " << r << " " << s << "\n";
  return 0;
}

### Prompt

Create a solution in cpp for 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 n, p, q, r, s;
long long b[2002][2002];
long long le[5000];
long long ri[5000];
int main() {
  scanf(" %I64d", &n);
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      scanf(" %I64d", &b[i][j]);
      le[i + j] += b[i][j];
      ri[n + i - j] += b[i][j];
    }
  }
  long long odd = -1, even = -1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      long long ans = le[i + j] + ri[n + i - j] - b[i][j];
      if ((i + j) & 1) {
        if (ans > odd) {
          odd = ans;
          p = i;
          q = j;
        }
      } else {
        if (ans > even) {
          even = ans;
          r = i;
          s = j;
        }
      }
    }
  }
  cout << even + odd << "\n" << p << " " << q << " " << r << " " << s << "\n";
  return 0;
}

```