#include <bits/stdc++.h>
using namespace std;
long long M[2005][2005];
long long S[2005][2005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, x1, y1, x2, y2;
  long long s1, s2;
  while (cin >> n) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) S[i][j] = 0, cin >> M[i][j];
    for (int r = 0; r < n; r++) {
      long long sum = 0;
      for (int i = r, j = 0; i < n; i++, j++) sum += M[i][j];
      for (int i = r, j = 0; i < n; i++, j++) S[i][j] += sum;
    }
    for (int c = 1; c < n; c++) {
      long long sum = 0;
      for (int i = 0, j = c; j < n; i++, j++) sum += M[i][j];
      for (int i = 0, j = c; j < n; i++, j++) S[i][j] += sum;
    }
    for (int r = 0; r < n; r++) {
      long long sum = 0;
      for (int i = r, j = n - 1; i < n; i++, j--) sum += M[i][j];
      for (int i = r, j = n - 1; i < n; i++, j--) S[i][j] += sum;
    }
    for (int c = 1; c < n; c++) {
      long long sum = 0;
      for (int i = 0, j = n - 1 - c; j > -1; i++, j--) sum += M[i][j];
      for (int i = 0, j = n - 1 - c; j > -1; i++, j--) S[i][j] += sum;
    }
    s1 = -1;
    s2 = -1;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        if ((i + j) % 2 == 0 && S[i][j] - M[i][j] > s1) {
          s1 = S[i][j] - M[i][j];
          x1 = i + 1;
          y1 = j + 1;
        } else if ((i + j) % 2 != 0 && S[i][j] - M[i][j] > s2) {
          s2 = S[i][j] - M[i][j];
          x2 = i + 1;
          y2 = j + 1;
        }
    cout << s1 + s2 << '\n';
    cout << x1 << " " << y1 << " " << x2 << " " << y2 << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long M[2005][2005];
long long S[2005][2005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, x1, y1, x2, y2;
  long long s1, s2;
  while (cin >> n) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) S[i][j] = 0, cin >> M[i][j];
    for (int r = 0; r < n; r++) {
      long long sum = 0;
      for (int i = r, j = 0; i < n; i++, j++) sum += M[i][j];
      for (int i = r, j = 0; i < n; i++, j++) S[i][j] += sum;
    }
    for (int c = 1; c < n; c++) {
      long long sum = 0;
      for (int i = 0, j = c; j < n; i++, j++) sum += M[i][j];
      for (int i = 0, j = c; j < n; i++, j++) S[i][j] += sum;
    }
    for (int r = 0; r < n; r++) {
      long long sum = 0;
      for (int i = r, j = n - 1; i < n; i++, j--) sum += M[i][j];
      for (int i = r, j = n - 1; i < n; i++, j--) S[i][j] += sum;
    }
    for (int c = 1; c < n; c++) {
      long long sum = 0;
      for (int i = 0, j = n - 1 - c; j > -1; i++, j--) sum += M[i][j];
      for (int i = 0, j = n - 1 - c; j > -1; i++, j--) S[i][j] += sum;
    }
    s1 = -1;
    s2 = -1;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        if ((i + j) % 2 == 0 && S[i][j] - M[i][j] > s1) {
          s1 = S[i][j] - M[i][j];
          x1 = i + 1;
          y1 = j + 1;
        } else if ((i + j) % 2 != 0 && S[i][j] - M[i][j] > s2) {
          s2 = S[i][j] - M[i][j];
          x2 = i + 1;
          y2 = j + 1;
        }
    cout << s1 + s2 << '\n';
    cout << x1 << " " << y1 << " " << x2 << " " << y2 << '\n';
  }
  return 0;
}

```