#include <bits/stdc++.h>
long long unsigned A[2222][2222], B[2222][2222];
int main() {
  unsigned n, i, j, k, x = 1, y = 1, X = 1, Y = 2;
  long long unsigned l = 0, r = 0, s;
  scanf("%u", &n);
  for (i = 0; i++ < n;)
    for (j = 0; j++ < n;) {
      scanf("%I64u", &s);
      A[i][j] = s + A[i - 1][j - 1];
      B[i][j] = s + B[i - 1][j + 1];
    }
  for (i = 0; i++ < n;)
    for (j = 0; j++ < n;) {
      s = A[i - 1][j - 1] - A[i][j];
      k = n - i;
      if (k > n - j) k = n - j;
      s += A[i + k][j + k];
      k = n - i;
      if (k > j - 1) k = j - 1;
      s += B[i + k][j - k];
      if ((i ^ j) & 1) {
        if (l < s) {
          X = i;
          Y = j;
          l = s;
        }
      } else if (r < s) {
        x = i;
        y = j;
        r = s;
      }
    }
  printf("%I64u\n", l + r);
  printf("%u %u %u %u\n", x, y, X, Y);
  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>
long long unsigned A[2222][2222], B[2222][2222];
int main() {
  unsigned n, i, j, k, x = 1, y = 1, X = 1, Y = 2;
  long long unsigned l = 0, r = 0, s;
  scanf("%u", &n);
  for (i = 0; i++ < n;)
    for (j = 0; j++ < n;) {
      scanf("%I64u", &s);
      A[i][j] = s + A[i - 1][j - 1];
      B[i][j] = s + B[i - 1][j + 1];
    }
  for (i = 0; i++ < n;)
    for (j = 0; j++ < n;) {
      s = A[i - 1][j - 1] - A[i][j];
      k = n - i;
      if (k > n - j) k = n - j;
      s += A[i + k][j + k];
      k = n - i;
      if (k > j - 1) k = j - 1;
      s += B[i + k][j - k];
      if ((i ^ j) & 1) {
        if (l < s) {
          X = i;
          Y = j;
          l = s;
        }
      } else if (r < s) {
        x = i;
        y = j;
        r = s;
      }
    }
  printf("%I64u\n", l + r);
  printf("%u %u %u %u\n", x, y, X, Y);
  return 0;
}

```