#include <bits/stdc++.h>
using namespace std;
int n, M[2000][2000];
long long sum1[2 * 2000], sum2[2 * 2000];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) scanf("%d", &M[i][j]);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      sum1[i + j] += M[i][j];
      sum2[i - j + n - 1] += M[i][j];
    }
  }
  long long best1 = -1, best2 = -1;
  int x1, y1, x2, y2;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      long long sum = sum1[i + j] + sum2[i - j + n - 1] - M[i][j];
      if (i % 2 == j % 2 && sum > best1) {
        best1 = sum;
        x1 = i + 1;
        y1 = j + 1;
      } else if (i % 2 != j % 2 && sum > best2) {
        best2 = sum;
        x2 = i + 1;
        y2 = j + 1;
      }
    }
  }
  printf("%I64d\n%d %d %d %d\n", best1 + best2, x1, y1, x2, y2);
  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 n, M[2000][2000];
long long sum1[2 * 2000], sum2[2 * 2000];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) scanf("%d", &M[i][j]);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      sum1[i + j] += M[i][j];
      sum2[i - j + n - 1] += M[i][j];
    }
  }
  long long best1 = -1, best2 = -1;
  int x1, y1, x2, y2;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      long long sum = sum1[i + j] + sum2[i - j + n - 1] - M[i][j];
      if (i % 2 == j % 2 && sum > best1) {
        best1 = sum;
        x1 = i + 1;
        y1 = j + 1;
      } else if (i % 2 != j % 2 && sum > best2) {
        best2 = sum;
        x2 = i + 1;
        y2 = j + 1;
      }
    }
  }
  printf("%I64d\n%d %d %d %d\n", best1 + best2, x1, y1, x2, y2);
  return 0;
}

```