#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, i, j, k, x, y, z;
  cin >> n;
  vector<int> v1(n, 0);
  vector<vector<int> > v;
  v.resize(n, v1);
  unordered_map<long long, long long> m1, m2;
  x = 1;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cin >> x;
      v[i][j] = x;
      m1[i + j] += x;
      m2[i - j] += x;
    }
  }
  long long maxodd = -1, maxeven = -1, x1, y1, x2, y2;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      z = m1[i + j] + m2[i - j] - v[i][j];
      if ((i + j) % 2 == 0) {
        if (z > maxeven) {
          maxeven = z;
          x1 = i + 1;
          y1 = j + 1;
        }
      } else {
        if (z > maxodd) {
          maxodd = z;
          x2 = i + 1;
          y2 = j + 1;
        }
      }
    }
  }
  cout << maxeven + maxodd << "\n";
  cout << x1 << " " << y1 << " " << x2 << " " << y2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t-- > 0) {
    solve();
  }
}

### Prompt

In Cpp, your task is to solve 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;
void solve() {
  long long n, i, j, k, x, y, z;
  cin >> n;
  vector<int> v1(n, 0);
  vector<vector<int> > v;
  v.resize(n, v1);
  unordered_map<long long, long long> m1, m2;
  x = 1;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cin >> x;
      v[i][j] = x;
      m1[i + j] += x;
      m2[i - j] += x;
    }
  }
  long long maxodd = -1, maxeven = -1, x1, y1, x2, y2;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      z = m1[i + j] + m2[i - j] - v[i][j];
      if ((i + j) % 2 == 0) {
        if (z > maxeven) {
          maxeven = z;
          x1 = i + 1;
          y1 = j + 1;
        }
      } else {
        if (z > maxodd) {
          maxodd = z;
          x2 = i + 1;
          y2 = j + 1;
        }
      }
    }
  }
  cout << maxeven + maxodd << "\n";
  cout << x1 << " " << y1 << " " << x2 << " " << y2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t-- > 0) {
    solve();
  }
}

```