#include <bits/stdc++.h>
using namespace std;
long long size, graph[2000][2000], tmp;
long long diag[2][4000], biggest[2];
int x[2], y[2], tmp2;
int main(int argc, const char* argv[]) {
  scanf("%I64d", &size);
  for (int i = 0; i < size; i++) {
    for (int k = 0; k < size; k++) {
      scanf("%I64d", &graph[i][k]);
      diag[0][i + k] += graph[i][k];
      diag[1][k - i + size] += graph[i][k];
    }
  }
  biggest[0] = -1, biggest[1] = -1;
  for (int i = 0; i < size; i++) {
    for (int j = 0; j < size; j++) {
      tmp = diag[0][i + j] + diag[1][j - i + size] - graph[i][j];
      tmp2 = (i + j) % 2;
      if (biggest[tmp2] < tmp) {
        biggest[tmp2] = tmp;
        x[tmp2] = i + 1;
        y[tmp2] = j + 1;
      }
    }
  }
  cout << biggest[0] + biggest[1] << "\n"
       << x[0] << " " << y[0] << " " << x[1] << " " << y[1];
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long size, graph[2000][2000], tmp;
long long diag[2][4000], biggest[2];
int x[2], y[2], tmp2;
int main(int argc, const char* argv[]) {
  scanf("%I64d", &size);
  for (int i = 0; i < size; i++) {
    for (int k = 0; k < size; k++) {
      scanf("%I64d", &graph[i][k]);
      diag[0][i + k] += graph[i][k];
      diag[1][k - i + size] += graph[i][k];
    }
  }
  biggest[0] = -1, biggest[1] = -1;
  for (int i = 0; i < size; i++) {
    for (int j = 0; j < size; j++) {
      tmp = diag[0][i + j] + diag[1][j - i + size] - graph[i][j];
      tmp2 = (i + j) % 2;
      if (biggest[tmp2] < tmp) {
        biggest[tmp2] = tmp;
        x[tmp2] = i + 1;
        y[tmp2] = j + 1;
      }
    }
  }
  cout << biggest[0] + biggest[1] << "\n"
       << x[0] << " " << y[0] << " " << x[1] << " " << y[1];
  return 0;
}

```