#include <bits/stdc++.h>
using namespace std;
const int N = 2005, MOD = 1e9 + 7, MAX = 1e5;
int n;
int a[N][N];
long long r[4 * N], l[4 * N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      scanf("%d", &a[i][j]);
      r[i + j] += a[i][j];
      l[i - j + n] += a[i][j];
    }
  }
  pair<long long, pair<int, int> > best[2];
  best[0].first = best[1].first = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((r[i + j] + l[i - j + n] - a[i][j]) > best[(i + j) % 2].first) {
        int wa = (i + j) % 2;
        best[wa].first = r[i + j] + l[i - j + n] - a[i][j];
        best[wa].second.first = i;
        best[wa].second.second = j;
      }
    }
  }
  printf("%lld\n%d %d %d %d", best[0].first + best[1].first,
         best[0].second.first, best[0].second.second, best[1].second.first,
         best[1].second.second);
  return 0;
}

### Prompt

Generate 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 namespace std;
const int N = 2005, MOD = 1e9 + 7, MAX = 1e5;
int n;
int a[N][N];
long long r[4 * N], l[4 * N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      scanf("%d", &a[i][j]);
      r[i + j] += a[i][j];
      l[i - j + n] += a[i][j];
    }
  }
  pair<long long, pair<int, int> > best[2];
  best[0].first = best[1].first = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((r[i + j] + l[i - j + n] - a[i][j]) > best[(i + j) % 2].first) {
        int wa = (i + j) % 2;
        best[wa].first = r[i + j] + l[i - j + n] - a[i][j];
        best[wa].second.first = i;
        best[wa].second.second = j;
      }
    }
  }
  printf("%lld\n%d %d %d %d", best[0].first + best[1].first,
         best[0].second.first, best[0].second.second, best[1].second.first,
         best[1].second.second);
  return 0;
}

```