#include <bits/stdc++.h>
using namespace std;
int n;
int mat[2005][2005];
unordered_map<int, long long> dijd;
long long dijl[2 * 2005];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = (0); (i) <= (n - 1); (i)++) {
    for (int j = (0); (j) <= (n - 1); (j)++) {
      cin >> mat[i][j];
      dijd[i - j] += mat[i][j];
      dijl[i + j] += mat[i][j];
    }
  }
  long long bela = 0;
  pair<int, int> id1, id2;
  id1 = {1, 1};
  id2 = {1, 2};
  long long crna = 0;
  for (int i = (0); (i) <= (n - 1); (i)++) {
    for (int j = (0); (j) <= (n - 1); (j)++) {
      long long c = dijd[i - j] + dijl[i + j] - mat[i][j];
      if ((i + j) % 2 == 0) {
        if (c > bela) {
          bela = c;
          id1 = {i, j};
        }
      } else {
        if (c > crna) {
          crna = c;
          id2 = {i, j};
        }
      }
    }
  }
  cout << bela + crna << endl;
  cout << id1.first + 1 << " " << id1.second + 1 << " " << id2.first + 1 << " "
       << id2.second + 1 << 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 n;
int mat[2005][2005];
unordered_map<int, long long> dijd;
long long dijl[2 * 2005];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = (0); (i) <= (n - 1); (i)++) {
    for (int j = (0); (j) <= (n - 1); (j)++) {
      cin >> mat[i][j];
      dijd[i - j] += mat[i][j];
      dijl[i + j] += mat[i][j];
    }
  }
  long long bela = 0;
  pair<int, int> id1, id2;
  id1 = {1, 1};
  id2 = {1, 2};
  long long crna = 0;
  for (int i = (0); (i) <= (n - 1); (i)++) {
    for (int j = (0); (j) <= (n - 1); (j)++) {
      long long c = dijd[i - j] + dijl[i + j] - mat[i][j];
      if ((i + j) % 2 == 0) {
        if (c > bela) {
          bela = c;
          id1 = {i, j};
        }
      } else {
        if (c > crna) {
          crna = c;
          id2 = {i, j};
        }
      }
    }
  }
  cout << bela + crna << endl;
  cout << id1.first + 1 << " " << id1.second + 1 << " " << id2.first + 1 << " "
       << id2.second + 1 << endl;
  return 0;
}

```