#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n][n], primary[n][n], second[n][n];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) cin >> a[i][j];
    }
    for (int i = 0; i < n; i++) {
      int x = i, y = 0;
      primary[x][y] = 0;
      while (x < n && y < n) {
        primary[i][0] += a[x][y];
        x++, y++;
      }
      x = i + 1, y = 1;
      while (x < n && y < n) {
        primary[x][y] = primary[x - 1][y - 1];
        x++, y++;
      }
    }
    for (int i = 1; i < n; i++) {
      int x = 0, y = i;
      primary[x][y] = 0;
      while (x < n && y < n) {
        primary[0][i] += a[x][y];
        x++, y++;
      }
      x = 1, y = i + 1;
      while (x < n && y < n) {
        primary[x][y] = primary[x - 1][y - 1];
        x++, y++;
      }
    }
    for (int i = 0; i < n; i++) {
      int x = i, y = 0;
      second[x][y] = 0;
      while (x >= 0 && y < n) {
        second[i][0] += a[x][y];
        x--, y++;
      }
      x = i - 1, y = 1;
      while (x >= 0 && y < n) {
        second[x][y] = second[x + 1][y - 1];
        x--, y++;
      }
    }
    for (int i = 1; i < n; i++) {
      int x = n - 1, y = i;
      second[x][y] = 0;
      while (x >= 0 && y < n) {
        second[n - 1][i] += a[x][y];
        x--, y++;
      }
      x = n - 2, y = i + 1;
      while (x >= 0 && y < n) {
        second[x][y] = second[x + 1][y - 1];
        x--, y++;
      }
    }
    long long even = 0, odd = 0, ex = 1, ey = 1, ox = 1, oy = 2;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if ((i + j + 2) % 2 == 0) {
          if (primary[i][j] + second[i][j] - a[i][j] > even) {
            even = primary[i][j] + second[i][j] - a[i][j];
            ex = i + 1, ey = j + 1;
          }
        } else {
          if (primary[i][j] + second[i][j] - a[i][j] > odd) {
            odd = primary[i][j] + second[i][j] - a[i][j];
            ox = i + 1, oy = j + 1;
          }
        }
      }
    }
    cout << odd + even << "\n";
    cout << ex << " " << ey << " " << ox << " " << oy << "\n";
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n][n], primary[n][n], second[n][n];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) cin >> a[i][j];
    }
    for (int i = 0; i < n; i++) {
      int x = i, y = 0;
      primary[x][y] = 0;
      while (x < n && y < n) {
        primary[i][0] += a[x][y];
        x++, y++;
      }
      x = i + 1, y = 1;
      while (x < n && y < n) {
        primary[x][y] = primary[x - 1][y - 1];
        x++, y++;
      }
    }
    for (int i = 1; i < n; i++) {
      int x = 0, y = i;
      primary[x][y] = 0;
      while (x < n && y < n) {
        primary[0][i] += a[x][y];
        x++, y++;
      }
      x = 1, y = i + 1;
      while (x < n && y < n) {
        primary[x][y] = primary[x - 1][y - 1];
        x++, y++;
      }
    }
    for (int i = 0; i < n; i++) {
      int x = i, y = 0;
      second[x][y] = 0;
      while (x >= 0 && y < n) {
        second[i][0] += a[x][y];
        x--, y++;
      }
      x = i - 1, y = 1;
      while (x >= 0 && y < n) {
        second[x][y] = second[x + 1][y - 1];
        x--, y++;
      }
    }
    for (int i = 1; i < n; i++) {
      int x = n - 1, y = i;
      second[x][y] = 0;
      while (x >= 0 && y < n) {
        second[n - 1][i] += a[x][y];
        x--, y++;
      }
      x = n - 2, y = i + 1;
      while (x >= 0 && y < n) {
        second[x][y] = second[x + 1][y - 1];
        x--, y++;
      }
    }
    long long even = 0, odd = 0, ex = 1, ey = 1, ox = 1, oy = 2;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if ((i + j + 2) % 2 == 0) {
          if (primary[i][j] + second[i][j] - a[i][j] > even) {
            even = primary[i][j] + second[i][j] - a[i][j];
            ex = i + 1, ey = j + 1;
          }
        } else {
          if (primary[i][j] + second[i][j] - a[i][j] > odd) {
            odd = primary[i][j] + second[i][j] - a[i][j];
            ox = i + 1, oy = j + 1;
          }
        }
      }
    }
    cout << odd + even << "\n";
    cout << ex << " " << ey << " " << ox << " " << oy << "\n";
  }
  return 0;
}

```