#include <bits/stdc++.h>
using namespace std;
int a[2010][2010];
long long int d1[4010];
long long int d2[4010];
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]);
      d1[i + j] += a[i][j];
      d2[n + i - j] += a[i][j];
    }
  }
  long long int odd = -1, even = -1;
  int ex = -1, ey = -1, ox = -1, oy = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (((i + j) & 1) == 0) {
        long long int val = d1[i + j] + d2[n + i - j] - a[i][j];
        if (val > even) {
          even = val;
          ex = i;
          ey = j;
        }
      } else {
        long long int val = d1[i + j] + d2[n + i - j] - a[i][j];
        if (val > odd) {
          odd = val;
          ox = i;
          oy = j;
        }
      }
    }
  }
  printf("%I64d\n", even + odd);
  printf("%d %d %d %d\n", ex + 1, ey + 1, ox + 1, oy + 1);
  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;
int a[2010][2010];
long long int d1[4010];
long long int d2[4010];
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]);
      d1[i + j] += a[i][j];
      d2[n + i - j] += a[i][j];
    }
  }
  long long int odd = -1, even = -1;
  int ex = -1, ey = -1, ox = -1, oy = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (((i + j) & 1) == 0) {
        long long int val = d1[i + j] + d2[n + i - j] - a[i][j];
        if (val > even) {
          even = val;
          ex = i;
          ey = j;
        }
      } else {
        long long int val = d1[i + j] + d2[n + i - j] - a[i][j];
        if (val > odd) {
          odd = val;
          ox = i;
          oy = j;
        }
      }
    }
  }
  printf("%I64d\n", even + odd);
  printf("%d %d %d %d\n", ex + 1, ey + 1, ox + 1, oy + 1);
  return 0;
}

```