#include <bits/stdc++.h>
using namespace std;
int a[2005][2005];
long long pri[4005], sec[4005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      scanf("%d", &a[i][j]);
      pri[i - j + n] += a[i][j];
      sec[i + j] += a[i][j];
    }
  }
  long long b = 0, w = 0, val;
  int pos_b[2] = {1, 1}, pos_w[2] = {1, 2};
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      val = pri[i - j + n] + sec[i + j] - a[i][j];
      if ((i + j) & 1) {
        if (val > w) w = val, pos_w[0] = i + 1, pos_w[1] = j + 1;
      } else {
        if (val > b) b = val, pos_b[0] = i + 1, pos_b[1] = j + 1;
      }
    }
  }
  cout << b + w << endl;
  cout << pos_b[0] << " " << pos_b[1] << " ";
  cout << pos_w[0] << " " << pos_w[1] << " ";
  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;
int a[2005][2005];
long long pri[4005], sec[4005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      scanf("%d", &a[i][j]);
      pri[i - j + n] += a[i][j];
      sec[i + j] += a[i][j];
    }
  }
  long long b = 0, w = 0, val;
  int pos_b[2] = {1, 1}, pos_w[2] = {1, 2};
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      val = pri[i - j + n] + sec[i + j] - a[i][j];
      if ((i + j) & 1) {
        if (val > w) w = val, pos_w[0] = i + 1, pos_w[1] = j + 1;
      } else {
        if (val > b) b = val, pos_b[0] = i + 1, pos_b[1] = j + 1;
      }
    }
  }
  cout << b + w << endl;
  cout << pos_b[0] << " " << pos_b[1] << " ";
  cout << pos_w[0] << " " << pos_w[1] << " ";
  return 0;
}

```