#include <bits/stdc++.h>
using namespace std;
int long long a[4007], b[4007];
signed main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int long long n;
  cin >> n;
  int long long vv[n + 1][n + 1];
  for (int long long i = 1; i <= n; i++) {
    for (int long long j = 1; j <= n; j++) {
      cin >> vv[i][j];
      a[i + j] += vv[i][j];
      b[i - j + n] += vv[i][j];
    }
  }
  int long long u = 0, v = 0, w = 0, x = 0, max1 = -1, max2 = -1;
  for (int long long i = 1; i <= n; i++) {
    for (int long long j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        int long long sum1 = a[i + j] + b[i - j + n] - vv[i][j];
        if (sum1 > max1) {
          u = i;
          v = j;
          max1 = sum1;
        }
      } else {
        int long long sum2 = a[i + j] + b[i - j + n] - vv[i][j];
        if (sum2 > max2) {
          w = i;
          x = j;
          max2 = sum2;
        }
      }
    }
  }
  cout << max1 + max2 << endl;
  if (u + v > w + x) {
    cout << w << " " << x << " " << u << " " << v << endl;
  } else {
    cout << u << " " << v << " " << w << " " << x << endl;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
using namespace std;
int long long a[4007], b[4007];
signed main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int long long n;
  cin >> n;
  int long long vv[n + 1][n + 1];
  for (int long long i = 1; i <= n; i++) {
    for (int long long j = 1; j <= n; j++) {
      cin >> vv[i][j];
      a[i + j] += vv[i][j];
      b[i - j + n] += vv[i][j];
    }
  }
  int long long u = 0, v = 0, w = 0, x = 0, max1 = -1, max2 = -1;
  for (int long long i = 1; i <= n; i++) {
    for (int long long j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        int long long sum1 = a[i + j] + b[i - j + n] - vv[i][j];
        if (sum1 > max1) {
          u = i;
          v = j;
          max1 = sum1;
        }
      } else {
        int long long sum2 = a[i + j] + b[i - j + n] - vv[i][j];
        if (sum2 > max2) {
          w = i;
          x = j;
          max2 = sum2;
        }
      }
    }
  }
  cout << max1 + max2 << endl;
  if (u + v > w + x) {
    cout << w << " " << x << " " << u << " " << v << endl;
  } else {
    cout << u << " " << v << " " << w << " " << x << endl;
  }
  return 0;
}

```