#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<long long int> > a(n, vector<long long int>(n));
  vector<long long> right(2 * n, 0);
  vector<long long> left(2 * n, 0);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      right[i - j + n] += a[i][j];
      left[i + j] += a[i][j];
    }
  long long wm = 0, bm = 0;
  int wmx = 1, wmy = 1, bmx = 1, bmy = 2;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      a[i][j] = right[i - j + n] - a[i][j] + left[i + j];
      if ((i + j) % 2 == 0) {
        if (a[i][j] > wm) {
          wm = a[i][j];
          wmx = i + 1;
          wmy = j + 1;
        }
      } else {
        if (a[i][j] > bm) {
          bm = a[i][j];
          bmx = i + 1;
          bmy = j + 1;
        }
      }
    }
  cout << wm + bm << '\n';
  cout << wmx << ' ' << wmy << ' ' << bmx << ' ' << bmy << ' ';
  return 0;
}

### 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;
int main() {
  cin.sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<long long int> > a(n, vector<long long int>(n));
  vector<long long> right(2 * n, 0);
  vector<long long> left(2 * n, 0);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      right[i - j + n] += a[i][j];
      left[i + j] += a[i][j];
    }
  long long wm = 0, bm = 0;
  int wmx = 1, wmy = 1, bmx = 1, bmy = 2;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      a[i][j] = right[i - j + n] - a[i][j] + left[i + j];
      if ((i + j) % 2 == 0) {
        if (a[i][j] > wm) {
          wm = a[i][j];
          wmx = i + 1;
          wmy = j + 1;
        }
      } else {
        if (a[i][j] > bm) {
          bm = a[i][j];
          bmx = i + 1;
          bmy = j + 1;
        }
      }
    }
  cout << wm + bm << '\n';
  cout << wmx << ' ' << wmy << ' ' << bmx << ' ' << bmy << ' ';
  return 0;
}

```