#include <bits/stdc++.h>
using namespace std;
const int N = 2003;
long long n, a[N][N], dp[2 * N], dp1[2 * N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      dp[i + j] += a[i][j];
      dp1[i - j + n] += a[i][j];
    }
  }
  long long y = -1, x = -1;
  pair<long long, long long> p, pp;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      long long k = dp[i + j] + dp1[i - j + n] - a[i][j];
      if ((i + j) & 1) {
        if (k > y) {
          y = k;
          p = {i, j};
        }
      } else {
        if (k > x) {
          x = k;
          pp = {i, j};
        }
      }
    }
  }
  cout << x + y << endl
       << p.first + 1 << " " << p.second + 1 << " " << pp.first + 1 << " "
       << pp.second + 1;
}

### Prompt

Create a solution in Cpp for 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;
const int N = 2003;
long long n, a[N][N], dp[2 * N], dp1[2 * N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      dp[i + j] += a[i][j];
      dp1[i - j + n] += a[i][j];
    }
  }
  long long y = -1, x = -1;
  pair<long long, long long> p, pp;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      long long k = dp[i + j] + dp1[i - j + n] - a[i][j];
      if ((i + j) & 1) {
        if (k > y) {
          y = k;
          p = {i, j};
        }
      } else {
        if (k > x) {
          x = k;
          pp = {i, j};
        }
      }
    }
  }
  cout << x + y << endl
       << p.first + 1 << " " << p.second + 1 << " " << pp.first + 1 << " "
       << pp.second + 1;
}

```