#include <bits/stdc++.h>
using namespace std;
long long ar[2005][2005];
long long arl[4010];
long long arr[4010];
long long ar1[2005][2005];
long long ar2[2005][2005];
long long far[2005][2005];
int vis[2005][2005];
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) cin >> ar[i][j];
  }
  int f = 1;
  for (int i = 1; i <= 2 * n - 1; i++) {
    if (i <= n) {
      int k = 1;
      int j = i;
      while (k <= i) {
        arl[f] += ar[k][j--];
        k++;
      }
    } else {
      int k = 1;
      int j = i - n;
      int j1 = n;
      while (k <= j) {
        arl[f] += ar[n - j + k][j1--];
        k++;
      }
    }
    f++;
  }
  int f1 = 1;
  for (int i = 1; i <= 2 * n - 1; i++) {
    if (i <= n) {
      int k = 1;
      int l = 1;
      while (k <= i) {
        arr[f1] += ar[n - i + k][l++];
        k++;
      }
    } else {
      int k = 1;
      int j = i - n;
      int l = 1;
      while (k <= j) {
        arr[f1] += ar[l++][n - j + k];
        k++;
      }
    }
    f1++;
  }
  int f2 = 1;
  for (int i = 1; i <= 2 * n - 1; i++) {
    if (i <= n) {
      int k = 1;
      int j = i;
      while (k <= i) {
        ar1[k][j--] = arl[f2];
        k++;
      }
    } else {
      int k = 1;
      int j = i - n;
      int j1 = n;
      while (k <= j) {
        ar1[n - j + k][j1--] = arl[f2];
        k++;
      }
    }
    f2++;
  }
  int f3 = 1;
  for (int i = 1; i <= 2 * n - 1; i++) {
    if (i <= n) {
      int k = 1;
      int l = 1;
      while (k <= i) {
        ar2[n - i + k][l++] = arr[f3];
        k++;
      }
    } else {
      int k = 1;
      int j = i - n;
      int l = 1;
      while (k <= j) {
        ar2[l++][n - j + k] = arr[f3];
        k++;
      }
    }
    f3++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      far[i][j] = ar1[i][j] + ar2[i][j] - ar[i][j];
    }
  }
  long long x = 0, y = 0;
  long long maxi = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (maxi <= far[i][j]) {
        maxi = far[i][j];
        x = i;
        y = j;
      }
    }
  }
  long long maxi2 = 0, x1 = 0, y1 = 0;
  if ((x + y) % 2 != 0) {
    for (int i = 1; i <= n; i++) {
      if (i % 2 == 0) {
        for (int j = 2; j <= n; j += 2) {
          if (far[i][j] >= maxi2) {
            maxi2 = far[i][j];
            x1 = i;
            y1 = j;
          }
        }
      } else {
        for (int j = 1; j <= n; j += 2) {
          if (far[i][j] >= maxi2) {
            maxi2 = far[i][j];
            x1 = i;
            y1 = j;
          }
        }
      }
    }
  } else {
    for (int i = 1; i <= n; i++) {
      if (i % 2 != 0) {
        for (int j = 2; j <= n; j += 2) {
          if (far[i][j] >= maxi2) {
            maxi2 = far[i][j];
            x1 = i;
            y1 = j;
          }
        }
      } else {
        for (int j = 1; j <= n; j += 2) {
          if (far[i][j] >= maxi2) {
            maxi2 = far[i][j];
            x1 = i;
            y1 = j;
          }
        }
      }
    }
  }
  cout << maxi + maxi2 << endl;
  cout << x << " " << y << " " << x1 << " " << y1;
  return 0;
}

### Prompt

Generate 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;
long long ar[2005][2005];
long long arl[4010];
long long arr[4010];
long long ar1[2005][2005];
long long ar2[2005][2005];
long long far[2005][2005];
int vis[2005][2005];
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) cin >> ar[i][j];
  }
  int f = 1;
  for (int i = 1; i <= 2 * n - 1; i++) {
    if (i <= n) {
      int k = 1;
      int j = i;
      while (k <= i) {
        arl[f] += ar[k][j--];
        k++;
      }
    } else {
      int k = 1;
      int j = i - n;
      int j1 = n;
      while (k <= j) {
        arl[f] += ar[n - j + k][j1--];
        k++;
      }
    }
    f++;
  }
  int f1 = 1;
  for (int i = 1; i <= 2 * n - 1; i++) {
    if (i <= n) {
      int k = 1;
      int l = 1;
      while (k <= i) {
        arr[f1] += ar[n - i + k][l++];
        k++;
      }
    } else {
      int k = 1;
      int j = i - n;
      int l = 1;
      while (k <= j) {
        arr[f1] += ar[l++][n - j + k];
        k++;
      }
    }
    f1++;
  }
  int f2 = 1;
  for (int i = 1; i <= 2 * n - 1; i++) {
    if (i <= n) {
      int k = 1;
      int j = i;
      while (k <= i) {
        ar1[k][j--] = arl[f2];
        k++;
      }
    } else {
      int k = 1;
      int j = i - n;
      int j1 = n;
      while (k <= j) {
        ar1[n - j + k][j1--] = arl[f2];
        k++;
      }
    }
    f2++;
  }
  int f3 = 1;
  for (int i = 1; i <= 2 * n - 1; i++) {
    if (i <= n) {
      int k = 1;
      int l = 1;
      while (k <= i) {
        ar2[n - i + k][l++] = arr[f3];
        k++;
      }
    } else {
      int k = 1;
      int j = i - n;
      int l = 1;
      while (k <= j) {
        ar2[l++][n - j + k] = arr[f3];
        k++;
      }
    }
    f3++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      far[i][j] = ar1[i][j] + ar2[i][j] - ar[i][j];
    }
  }
  long long x = 0, y = 0;
  long long maxi = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (maxi <= far[i][j]) {
        maxi = far[i][j];
        x = i;
        y = j;
      }
    }
  }
  long long maxi2 = 0, x1 = 0, y1 = 0;
  if ((x + y) % 2 != 0) {
    for (int i = 1; i <= n; i++) {
      if (i % 2 == 0) {
        for (int j = 2; j <= n; j += 2) {
          if (far[i][j] >= maxi2) {
            maxi2 = far[i][j];
            x1 = i;
            y1 = j;
          }
        }
      } else {
        for (int j = 1; j <= n; j += 2) {
          if (far[i][j] >= maxi2) {
            maxi2 = far[i][j];
            x1 = i;
            y1 = j;
          }
        }
      }
    }
  } else {
    for (int i = 1; i <= n; i++) {
      if (i % 2 != 0) {
        for (int j = 2; j <= n; j += 2) {
          if (far[i][j] >= maxi2) {
            maxi2 = far[i][j];
            x1 = i;
            y1 = j;
          }
        }
      } else {
        for (int j = 1; j <= n; j += 2) {
          if (far[i][j] >= maxi2) {
            maxi2 = far[i][j];
            x1 = i;
            y1 = j;
          }
        }
      }
    }
  }
  cout << maxi + maxi2 << endl;
  cout << x << " " << y << " " << x1 << " " << y1;
  return 0;
}

```