#include <bits/stdc++.h>
using vi = std::vector<int64_t>;
using vvi = std::vector<vi>;
struct N {
  int x;
  int y;
  int64_t sum;
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  vvi a(n, vi(n));
  vvi m(n, vi(n));
  int diags = 2 * n - 1;
  vi d1(diags), d2(diags);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      std::cin >> a[i][j];
      d1[i + j] += a[i][j];
      d2[i - j + n - 1] += a[i][j];
    }
  }
  std::vector<N> best({{0, 0, 0}, {0, 1, 0}});
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      int64_t sum = d1[i + j] - a[i][j] + d2[i - j + n - 1];
      int b = (i + j) % 2;
      if (sum > best[b].sum) {
        best[b] = {i, j, sum};
      }
    }
  }
  std::cout << best[0].sum + best[1].sum << std::endl;
  std::cout << best[0].x + 1 << " " << best[0].y + 1 << " " << best[1].x + 1
            << " " << best[1].y + 1 << std::endl;
}

### Prompt

Your challenge is to write 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 vi = std::vector<int64_t>;
using vvi = std::vector<vi>;
struct N {
  int x;
  int y;
  int64_t sum;
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  vvi a(n, vi(n));
  vvi m(n, vi(n));
  int diags = 2 * n - 1;
  vi d1(diags), d2(diags);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      std::cin >> a[i][j];
      d1[i + j] += a[i][j];
      d2[i - j + n - 1] += a[i][j];
    }
  }
  std::vector<N> best({{0, 0, 0}, {0, 1, 0}});
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      int64_t sum = d1[i + j] - a[i][j] + d2[i - j + n - 1];
      int b = (i + j) % 2;
      if (sum > best[b].sum) {
        best[b] = {i, j, sum};
      }
    }
  }
  std::cout << best[0].sum + best[1].sum << std::endl;
  std::cout << best[0].x + 1 << " " << best[0].y + 1 << " " << best[1].x + 1
            << " " << best[1].y + 1 << std::endl;
}

```