#include <bits/stdc++.h>
using namespace std;
const int NMax = 2000 + 10;
int N;
long long a[NMax][NMax];
long long prinDiag[5 * NMax], secDiag[5 * NMax];
int main() {
  ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= N; ++j) {
      cin >> a[i][j];
      int index1 = NMax + i - j, index2 = NMax + i + j;
      prinDiag[index1] += a[i][j];
      secDiag[index2] += a[i][j];
    }
  }
  long long max1, x1, y1, max2, x2, y2, val;
  max1 = max2 = -1;
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= N; ++j) {
      int index1 = NMax + i - j, index2 = NMax + i + j;
      val = prinDiag[index1] + secDiag[index2] - a[i][j];
      if ((i + j) % 2 == 0 && max1 < val) {
        max1 = val;
        x1 = i;
        y1 = j;
      } else if ((i + j) % 2 == 1 && max2 < val) {
        max2 = val;
        x2 = i;
        y2 = j;
      }
    }
  }
  cout << max1 + max2 << '\n' << x1 << ' ' << y1 << ' ' << x2 << ' ' << y2;
  return 0;
}

### 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;
const int NMax = 2000 + 10;
int N;
long long a[NMax][NMax];
long long prinDiag[5 * NMax], secDiag[5 * NMax];
int main() {
  ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= N; ++j) {
      cin >> a[i][j];
      int index1 = NMax + i - j, index2 = NMax + i + j;
      prinDiag[index1] += a[i][j];
      secDiag[index2] += a[i][j];
    }
  }
  long long max1, x1, y1, max2, x2, y2, val;
  max1 = max2 = -1;
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= N; ++j) {
      int index1 = NMax + i - j, index2 = NMax + i + j;
      val = prinDiag[index1] + secDiag[index2] - a[i][j];
      if ((i + j) % 2 == 0 && max1 < val) {
        max1 = val;
        x1 = i;
        y1 = j;
      } else if ((i + j) % 2 == 1 && max2 < val) {
        max2 = val;
        x2 = i;
        y2 = j;
      }
    }
  }
  cout << max1 + max2 << '\n' << x1 << ' ' << y1 << ' ' << x2 << ' ' << y2;
  return 0;
}

```