#include <bits/stdc++.h>
using namespace std;
int FASTBUFFER;
const int N = 2005;
long long n, a[N][N], ans[2], ansx[2], ansy[2];
long long v1[N << 2], v2[N << 2];
int main() {
  scanf("%I64d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      scanf("%I64d", &a[i][j]);
      v1[i + j] += a[i][j];
      v2[i - j + 2000] += a[i][j];
    }
  }
  ans[0] = ans[1] = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      int x = i + j & 1;
      long long s = v1[i + j] + v2[i - j + 2000] - a[i][j];
      if (s > ans[x]) {
        ans[x] = s;
        ansx[x] = i;
        ansy[x] = j;
      }
    }
  }
  printf("%I64d\n", ans[0] + ans[1]);
  printf("%I64d %I64d %I64d %I64d\n", ansx[0], ansy[0], ansx[1], ansy[1]);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 FASTBUFFER;
const int N = 2005;
long long n, a[N][N], ans[2], ansx[2], ansy[2];
long long v1[N << 2], v2[N << 2];
int main() {
  scanf("%I64d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      scanf("%I64d", &a[i][j]);
      v1[i + j] += a[i][j];
      v2[i - j + 2000] += a[i][j];
    }
  }
  ans[0] = ans[1] = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      int x = i + j & 1;
      long long s = v1[i + j] + v2[i - j + 2000] - a[i][j];
      if (s > ans[x]) {
        ans[x] = s;
        ansx[x] = i;
        ansy[x] = j;
      }
    }
  }
  printf("%I64d\n", ans[0] + ans[1]);
  printf("%I64d %I64d %I64d %I64d\n", ansx[0], ansy[0], ansx[1], ansy[1]);
  return 0;
}

```