#include <bits/stdc++.h>
using namespace std;
void boost() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
unsigned long long a[2001][2001];
unsigned long long d[8001];
unsigned long long d1[8001];
int main() {
  boost();
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      d[i + j] += a[i][j];
      if (i - j < 0) {
        d1[(i - j) + 2 * n] += a[i][j];
      } else {
        d1[i - j] += a[i][j];
      }
    }
  }
  pair<unsigned long long, pair<int, int>> mx = make_pair(0, make_pair(0, 0));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        unsigned long long k = d[i + j];
        if (i - j < 0) {
          k += d1[(i - j) + 2 * n];
        } else {
          k += d1[i - j];
        }
        k -= a[i][j];
        if (mx.first < k) {
          mx.first = k;
          mx.second.first = i;
          mx.second.second = j;
        }
      }
    }
  }
  pair<unsigned long long, pair<int, int>> mx1 = make_pair(0, make_pair(0, 0));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((i + j) % 2) {
        unsigned long long k = d[i + j];
        if (i - j < 0) {
          k += d1[(i - j) + 2 * n];
        } else {
          k += d1[i - j];
        }
        k -= a[i][j];
        if (mx1.first < k) {
          mx1.first = k;
          mx1.second.first = i;
          mx1.second.second = j;
        }
      }
    }
  }
  if (mx.first + mx1.first == 0) {
    cout << 0 << endl << "1 1 1 2";
    return 0;
  }
  cout << mx.first + mx1.first << endl
       << mx.second.first << " " << mx.second.second << " " << mx1.second.first
       << " " << mx1.second.second;
}

### 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>
using namespace std;
void boost() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
unsigned long long a[2001][2001];
unsigned long long d[8001];
unsigned long long d1[8001];
int main() {
  boost();
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      d[i + j] += a[i][j];
      if (i - j < 0) {
        d1[(i - j) + 2 * n] += a[i][j];
      } else {
        d1[i - j] += a[i][j];
      }
    }
  }
  pair<unsigned long long, pair<int, int>> mx = make_pair(0, make_pair(0, 0));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        unsigned long long k = d[i + j];
        if (i - j < 0) {
          k += d1[(i - j) + 2 * n];
        } else {
          k += d1[i - j];
        }
        k -= a[i][j];
        if (mx.first < k) {
          mx.first = k;
          mx.second.first = i;
          mx.second.second = j;
        }
      }
    }
  }
  pair<unsigned long long, pair<int, int>> mx1 = make_pair(0, make_pair(0, 0));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((i + j) % 2) {
        unsigned long long k = d[i + j];
        if (i - j < 0) {
          k += d1[(i - j) + 2 * n];
        } else {
          k += d1[i - j];
        }
        k -= a[i][j];
        if (mx1.first < k) {
          mx1.first = k;
          mx1.second.first = i;
          mx1.second.second = j;
        }
      }
    }
  }
  if (mx.first + mx1.first == 0) {
    cout << 0 << endl << "1 1 1 2";
    return 0;
  }
  cout << mx.first + mx1.first << endl
       << mx.second.first << " " << mx.second.second << " " << mx1.second.first
       << " " << mx1.second.second;
}

```