#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n;
  cin >> n;
  long long a[n + 3][n + 3];
  long long l[2 * n + 10], r[2 * n + 10];
  memset(l, 0, sizeof l);
  memset(r, 0, sizeof r);
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) {
      scanf("%lld", &a[i][j]);
      r[i + j] += a[i][j];
      l[n - 1 - i + j] += a[i][j];
    }
  long long ex = -1, ey = -1, ox = -1, oy = -1, es = -1, os = -1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      long long ss = r[i + j] + l[n - 1 - i + j] - a[i][j];
      if ((i + j) & 1) {
        if (ss > os) {
          os = ss;
          ox = i + 1;
          oy = j + 1;
        }
      } else {
        if (ss > es) {
          es = ss;
          ex = i + 1;
          ey = j + 1;
        }
      }
    }
  }
  printf("%lld\n", os + es);
  cout << ex << " " << ey << " " << ox << " " << oy << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 main() {
  long long n;
  cin >> n;
  long long a[n + 3][n + 3];
  long long l[2 * n + 10], r[2 * n + 10];
  memset(l, 0, sizeof l);
  memset(r, 0, sizeof r);
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) {
      scanf("%lld", &a[i][j]);
      r[i + j] += a[i][j];
      l[n - 1 - i + j] += a[i][j];
    }
  long long ex = -1, ey = -1, ox = -1, oy = -1, es = -1, os = -1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      long long ss = r[i + j] + l[n - 1 - i + j] - a[i][j];
      if ((i + j) & 1) {
        if (ss > os) {
          os = ss;
          ox = i + 1;
          oy = j + 1;
        }
      } else {
        if (ss > es) {
          es = ss;
          ex = i + 1;
          ey = j + 1;
        }
      }
    }
  }
  printf("%lld\n", os + es);
  cout << ex << " " << ey << " " << ox << " " << oy << endl;
  return 0;
}

```