#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int N = 4 * 1e3 + 5;
long long diag1[N], diag2[N];
int board[N][N];
int main() {
  int n, x;
  scanf("%d", &n);
  for (int r = (0); r < (n); ++r) {
    for (int c = (0); c < (n); ++c) {
      scanf("%d", &x);
      board[r][c] = x;
      diag1[r + c] += x;
      diag2[n - r + c] += x;
    }
  }
  long long score[2] = {-1, -1};
  pair<int, int> bishop[2] = {};
  for (int r = (0); r < (n); ++r) {
    for (int c = (0); c < (n); ++c) {
      if (diag1[r + c] + diag2[n - r + c] - board[r][c] > score[(r + c) % 2]) {
        score[(r + c) % 2] = diag1[r + c] + diag2[n - r + c] - board[r][c];
        bishop[(r + c) % 2] = {r, c};
      }
    }
  }
  cout << score[0] + score[1] << endl;
  printf("%d %d %d %d\n", bishop[0].first + 1, bishop[0].second + 1,
         bishop[1].first + 1, bishop[1].second + 1);
  return 0;
}

### 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;
const int INF = 1e9 + 7;
const int N = 4 * 1e3 + 5;
long long diag1[N], diag2[N];
int board[N][N];
int main() {
  int n, x;
  scanf("%d", &n);
  for (int r = (0); r < (n); ++r) {
    for (int c = (0); c < (n); ++c) {
      scanf("%d", &x);
      board[r][c] = x;
      diag1[r + c] += x;
      diag2[n - r + c] += x;
    }
  }
  long long score[2] = {-1, -1};
  pair<int, int> bishop[2] = {};
  for (int r = (0); r < (n); ++r) {
    for (int c = (0); c < (n); ++c) {
      if (diag1[r + c] + diag2[n - r + c] - board[r][c] > score[(r + c) % 2]) {
        score[(r + c) % 2] = diag1[r + c] + diag2[n - r + c] - board[r][c];
        bishop[(r + c) % 2] = {r, c};
      }
    }
  }
  cout << score[0] + score[1] << endl;
  printf("%d %d %d %d\n", bishop[0].first + 1, bishop[0].second + 1,
         bishop[1].first + 1, bishop[1].second + 1);
  return 0;
}

```