#include <bits/stdc++.h>
using namespace std;
int n;
long long mat[2020][2020], pos[4040], neg[4040];
int main() {
  scanf("%d", &n);
  memset(pos, 0, sizeof(pos));
  memset(neg, 0, sizeof(neg));
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      scanf("%I64d", &mat[i][j]);
      pos[i + j - 2] += mat[i][j];
      neg[i - j + n - 1] += mat[i][j];
    }
  }
  pair<int, int> best[2];
  long long val[2] = {-1LL, -1LL};
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      long long sum = pos[i + j - 2] + neg[i - j + n - 1] - mat[i][j];
      ;
      if (val[0] < sum) {
        val[0] = sum;
        best[0] = pair<int, int>(i, j);
      }
      if (j < n) {
        ++j;
        sum = pos[i + j - 2] + neg[i - j + n - 1] - mat[i][j];
        if (val[1] < sum) {
          val[1] = sum;
          best[1] = pair<int, int>(i, j);
        }
      }
    }
    swap(best[0], best[1]);
    swap(val[0], val[1]);
  }
  printf("%I64d\n", val[0] + val[1]);
  printf("%d %d %d %d\n", best[0].first, best[0].second, best[1].first,
         best[1].second);
  return 0;
}

### Prompt

Please formulate 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 n;
long long mat[2020][2020], pos[4040], neg[4040];
int main() {
  scanf("%d", &n);
  memset(pos, 0, sizeof(pos));
  memset(neg, 0, sizeof(neg));
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      scanf("%I64d", &mat[i][j]);
      pos[i + j - 2] += mat[i][j];
      neg[i - j + n - 1] += mat[i][j];
    }
  }
  pair<int, int> best[2];
  long long val[2] = {-1LL, -1LL};
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      long long sum = pos[i + j - 2] + neg[i - j + n - 1] - mat[i][j];
      ;
      if (val[0] < sum) {
        val[0] = sum;
        best[0] = pair<int, int>(i, j);
      }
      if (j < n) {
        ++j;
        sum = pos[i + j - 2] + neg[i - j + n - 1] - mat[i][j];
        if (val[1] < sum) {
          val[1] = sum;
          best[1] = pair<int, int>(i, j);
        }
      }
    }
    swap(best[0], best[1]);
    swap(val[0], val[1]);
  }
  printf("%I64d\n", val[0] + val[1]);
  printf("%d %d %d %d\n", best[0].first, best[0].second, best[1].first,
         best[1].second);
  return 0;
}

```