#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long a[n][n];
  long long x1, x2, y1, y2;
  map<long long, long long> m1;
  map<long long, long long> m2;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      cin >> a[i][j];
      m1[i + j] += a[i][j];
      m2[i - j] += a[i][j];
    }
  }
  long long ma1 = -1000000;
  for (long long i = 0; i < n; i++) {
    for (long long j = i % 2; j < n; j += 2)
      if (m1[i + j] + m2[i - j] - a[i][j] >= ma1) {
        ma1 = m1[i + j] + m2[i - j] - a[i][j];
        x1 = i + 1;
        y1 = j + 1;
      }
  }
  long long ma2 = -10000;
  for (long long i = 0; i < n; i++) {
    for (long long j = (i + 1) % 2; j < n; j += 2)
      if (m1[i + j] + m2[i - j] - a[i][j] >= ma2) {
        ma2 = m1[i + j] + m2[i - j] - a[i][j];
        x2 = i + 1;
        y2 = j + 1;
      }
  }
  cout << ma1 + ma2 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2;
}

### Prompt

Develop a solution in CPP to the problem described below:

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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long a[n][n];
  long long x1, x2, y1, y2;
  map<long long, long long> m1;
  map<long long, long long> m2;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      cin >> a[i][j];
      m1[i + j] += a[i][j];
      m2[i - j] += a[i][j];
    }
  }
  long long ma1 = -1000000;
  for (long long i = 0; i < n; i++) {
    for (long long j = i % 2; j < n; j += 2)
      if (m1[i + j] + m2[i - j] - a[i][j] >= ma1) {
        ma1 = m1[i + j] + m2[i - j] - a[i][j];
        x1 = i + 1;
        y1 = j + 1;
      }
  }
  long long ma2 = -10000;
  for (long long i = 0; i < n; i++) {
    for (long long j = (i + 1) % 2; j < n; j += 2)
      if (m1[i + j] + m2[i - j] - a[i][j] >= ma2) {
        ma2 = m1[i + j] + m2[i - j] - a[i][j];
        x2 = i + 1;
        y2 = j + 1;
      }
  }
  cout << ma1 + ma2 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2;
}

```