#include <bits/stdc++.h>
using namespace std;
int n, a[2001][2001];
long long lsum[4001], rsum[4001];
int xa, xb, ya, yb;
long long maxa, maxb;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      scanf("%d", &a[i][j]), lsum[i + j - 1] += a[i][j],
          rsum[i - j + n] += a[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      long long ans = lsum[i + j - 1] + rsum[i - j + n] - a[i][j];
      if ((i + j) & 1) {
        if (ans >= maxb) maxb = ans, xb = i, yb = j;
      } else if (ans >= maxa)
        maxa = ans, xa = i, ya = j;
    }
  printf("%lld\n%d %d %d %d", maxa + maxb, xa, ya, xb, yb);
  return 0;
}

### Prompt

Generate 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, a[2001][2001];
long long lsum[4001], rsum[4001];
int xa, xb, ya, yb;
long long maxa, maxb;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      scanf("%d", &a[i][j]), lsum[i + j - 1] += a[i][j],
          rsum[i - j + n] += a[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      long long ans = lsum[i + j - 1] + rsum[i - j + n] - a[i][j];
      if ((i + j) & 1) {
        if (ans >= maxb) maxb = ans, xb = i, yb = j;
      } else if (ans >= maxa)
        maxa = ans, xa = i, ya = j;
    }
  printf("%lld\n%d %d %d %d", maxa + maxb, xa, ya, xb, yb);
  return 0;
}

```