#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  scanf("%d", &n);
  int a[n][n];
  long long l[n][n], r[n][n];
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      scanf("%d", &a[i][j]);
      l[i][j] = r[i][j] = a[i][j];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i > 0 && j > 0) r[i][j] += r[i - 1][j - 1];
      if (i > 0 && j < n - 1) l[i][j] += l[i - 1][j + 1];
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    for (int j = n - 1; j >= 0; --j) {
      if (i < n - 1 && j < n - 1) r[i][j] = r[i + 1][j + 1];
      if (i < n - 1 && j > 0) l[i][j] = l[i + 1][j - 1];
    }
  }
  long long max1 = -1;
  int x1, y1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (l[i][j] + r[i][j] - a[i][j] > max1) {
        max1 = l[i][j] + r[i][j] - a[i][j];
        x1 = i;
        y1 = j;
      }
    }
  }
  long long max2 = -1;
  int x2, y2;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      int d = abs(i - x1) + abs(j - y1);
      if ((d & 1) && l[i][j] + r[i][j] - a[i][j] > max2) {
        max2 = l[i][j] + r[i][j] - a[i][j];
        x2 = i;
        y2 = j;
      }
    }
  }
  printf("%I64d\n", max1 + max2);
  printf("%d %d %d %d\n", x1 + 1, y1 + 1, x2 + 1, y2 + 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 main() {
  ios_base::sync_with_stdio(false);
  int n;
  scanf("%d", &n);
  int a[n][n];
  long long l[n][n], r[n][n];
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      scanf("%d", &a[i][j]);
      l[i][j] = r[i][j] = a[i][j];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i > 0 && j > 0) r[i][j] += r[i - 1][j - 1];
      if (i > 0 && j < n - 1) l[i][j] += l[i - 1][j + 1];
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    for (int j = n - 1; j >= 0; --j) {
      if (i < n - 1 && j < n - 1) r[i][j] = r[i + 1][j + 1];
      if (i < n - 1 && j > 0) l[i][j] = l[i + 1][j - 1];
    }
  }
  long long max1 = -1;
  int x1, y1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (l[i][j] + r[i][j] - a[i][j] > max1) {
        max1 = l[i][j] + r[i][j] - a[i][j];
        x1 = i;
        y1 = j;
      }
    }
  }
  long long max2 = -1;
  int x2, y2;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      int d = abs(i - x1) + abs(j - y1);
      if ((d & 1) && l[i][j] + r[i][j] - a[i][j] > max2) {
        max2 = l[i][j] + r[i][j] - a[i][j];
        x2 = i;
        y2 = j;
      }
    }
  }
  printf("%I64d\n", max1 + max2);
  printf("%d %d %d %d\n", x1 + 1, y1 + 1, x2 + 1, y2 + 1);
  return 0;
}

```