#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  long long arr[n + 1][n + 1];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> arr[i][j];
    }
  }
  long long m1 = 0, x1, y1, x2, y2, m2 = 0;
  unordered_map<int, long long> mp1, mp2;
  for (int i = 2; i <= n + 1; i++) {
    for (int y = 1; i - y >= 1; y++) {
      mp1[i] += arr[i - y][y];
    }
  }
  for (int i = n + 2; i <= 2 * n; i++) {
    for (int x = n; i - x <= n; x--) mp1[i] += arr[x][i - x];
  }
  for (int i = 0; i <= n - 1; i++) {
    for (int y = 1; i + y <= n; y++) mp2[i] += arr[i + y][y];
  }
  for (int i = -1; i >= 1 - n; i--) {
    for (int x = 1; x - i <= n; x++) mp2[i] += arr[x][x - i];
  }
  int p = 0;
  for (int i = 1; i <= n; i++) {
    int l;
    if (p == 0) {
      l = 0;
      p = 1;
    } else {
      l = 1;
      p = 0;
    }
    for (int j = 1; j <= n; j++) {
      if (l % 2 == 0) {
        if (mp1[i + j] + mp2[i - j] - arr[i][j] >= m1) {
          x1 = i;
          y1 = j;
          m1 = mp1[i + j] + mp2[i - j] - arr[i][j];
        }
      } else {
        if (mp1[i + j] + mp2[i - j] - arr[i][j] >= m2) {
          x2 = i;
          y2 = j;
          m2 = mp1[i + j] + mp2[i - j] - arr[i][j];
        }
      }
      l++;
    }
  }
  cout << (m1 + m2) << endl << 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(NULL);
  int n;
  cin >> n;
  long long arr[n + 1][n + 1];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> arr[i][j];
    }
  }
  long long m1 = 0, x1, y1, x2, y2, m2 = 0;
  unordered_map<int, long long> mp1, mp2;
  for (int i = 2; i <= n + 1; i++) {
    for (int y = 1; i - y >= 1; y++) {
      mp1[i] += arr[i - y][y];
    }
  }
  for (int i = n + 2; i <= 2 * n; i++) {
    for (int x = n; i - x <= n; x--) mp1[i] += arr[x][i - x];
  }
  for (int i = 0; i <= n - 1; i++) {
    for (int y = 1; i + y <= n; y++) mp2[i] += arr[i + y][y];
  }
  for (int i = -1; i >= 1 - n; i--) {
    for (int x = 1; x - i <= n; x++) mp2[i] += arr[x][x - i];
  }
  int p = 0;
  for (int i = 1; i <= n; i++) {
    int l;
    if (p == 0) {
      l = 0;
      p = 1;
    } else {
      l = 1;
      p = 0;
    }
    for (int j = 1; j <= n; j++) {
      if (l % 2 == 0) {
        if (mp1[i + j] + mp2[i - j] - arr[i][j] >= m1) {
          x1 = i;
          y1 = j;
          m1 = mp1[i + j] + mp2[i - j] - arr[i][j];
        }
      } else {
        if (mp1[i + j] + mp2[i - j] - arr[i][j] >= m2) {
          x2 = i;
          y2 = j;
          m2 = mp1[i + j] + mp2[i - j] - arr[i][j];
        }
      }
      l++;
    }
  }
  cout << (m1 + m2) << endl << x1 << ' ' << y1 << ' ' << x2 << ' ' << y2;
}

```