#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
long long d1[N * 2], d2[N * 2], ans[2];
pair<int, int> cells[2];
int a[N][N];
int main() {
  int n;
  scanf("%d", &n);
  int m = n * 2;
  for (int i = 0; i < m; i++) d1[i] = d2[i] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      scanf("%d", &a[i][j]);
      d1[i + j] += a[i][j];
      d2[n - i + j - 1] += a[i][j];
    }
  ans[0] = ans[1] = 0;
  cells[0] = make_pair(1, 1);
  cells[1] = make_pair(2, 1);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      long long sum = d1[i + j] + d2[n - i + j - 1] - a[i][j];
      if ((i + j) & 1) {
        if (sum > ans[0]) {
          ans[0] = sum;
          cells[0] = make_pair(i + 1, j + 1);
        }
      } else {
        if (sum > ans[1]) {
          ans[1] = sum;
          cells[1] = make_pair(i + 1, j + 1);
        }
      }
    }
  printf("%I64d\n", ans[0] + ans[1]);
  printf("%d %d %d %d\n", cells[0].first, cells[0].second, cells[1].first,
         cells[1].second);
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int N = 2005;
long long d1[N * 2], d2[N * 2], ans[2];
pair<int, int> cells[2];
int a[N][N];
int main() {
  int n;
  scanf("%d", &n);
  int m = n * 2;
  for (int i = 0; i < m; i++) d1[i] = d2[i] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      scanf("%d", &a[i][j]);
      d1[i + j] += a[i][j];
      d2[n - i + j - 1] += a[i][j];
    }
  ans[0] = ans[1] = 0;
  cells[0] = make_pair(1, 1);
  cells[1] = make_pair(2, 1);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      long long sum = d1[i + j] + d2[n - i + j - 1] - a[i][j];
      if ((i + j) & 1) {
        if (sum > ans[0]) {
          ans[0] = sum;
          cells[0] = make_pair(i + 1, j + 1);
        }
      } else {
        if (sum > ans[1]) {
          ans[1] = sum;
          cells[1] = make_pair(i + 1, j + 1);
        }
      }
    }
  printf("%I64d\n", ans[0] + ans[1]);
  printf("%d %d %d %d\n", cells[0].first, cells[0].second, cells[1].first,
         cells[1].second);
  return 0;
}

```