#include <bits/stdc++.h>
using namespace std;
struct item {
  long long l, r, x;
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int n, xb = 1, yb = 1, xw = 1, yw = 2;
  cin >> n;
  long long d1[n + n + 5], d2[n + n + 5], t[n + 1][n + 1];
  memset(d1, 0, sizeof d1), memset(d2, 0, sizeof d2);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      cin >> t[i][j];
      d1[i + j] += t[i][j];
      d2[i - j + n] += t[i][j];
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) t[i][j] = d1[i + j] + d2[i - j + n] - t[i][j];
  long long mxb = 0, mxw = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        if (t[i][j] > mxb) {
          mxb = t[i][j];
          xb = i;
          yb = j;
        }
      } else {
        if (t[i][j] > mxw) {
          mxw = t[i][j];
          xw = i;
          yw = j;
        }
      }
    }
  }
  cout << mxb + mxw << '\n';
  cout << xb << ' ' << yb << ' ' << xw << ' ' << yw;
  return 0;
}

### Prompt

Your task is to create 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;
struct item {
  long long l, r, x;
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int n, xb = 1, yb = 1, xw = 1, yw = 2;
  cin >> n;
  long long d1[n + n + 5], d2[n + n + 5], t[n + 1][n + 1];
  memset(d1, 0, sizeof d1), memset(d2, 0, sizeof d2);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      cin >> t[i][j];
      d1[i + j] += t[i][j];
      d2[i - j + n] += t[i][j];
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) t[i][j] = d1[i + j] + d2[i - j + n] - t[i][j];
  long long mxb = 0, mxw = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        if (t[i][j] > mxb) {
          mxb = t[i][j];
          xb = i;
          yb = j;
        }
      } else {
        if (t[i][j] > mxw) {
          mxw = t[i][j];
          xw = i;
          yw = j;
        }
      }
    }
  }
  cout << mxb + mxw << '\n';
  cout << xb << ' ' << yb << ' ' << xw << ' ' << yw;
  return 0;
}

```