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
#include <bits/stdc++.h>
using namespace std;
int n, temp, arr[2000 + 5][2000 + 5];
long long rl[2 * 2000 + 5], lr[2 * 2000 + 5];
pair<long long, pair<int, int> > xaml[4], xamr[4];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      scanf("%d", &arr[i][j]);
      rl[i + j] += arr[i][j];
      lr[n + i - j - 1] += arr[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i + j) % 2) {
        if (xaml[1].first <= rl[i + j] + lr[n + i - j - 1] - arr[i][j]) {
          xaml[1].first = rl[i + j] + lr[n + i - j - 1] - arr[i][j];
          xaml[1].second = make_pair(i + 1, j + 1);
          if (xaml[0].first <= xaml[1].first) swap(xaml[0], xaml[1]);
        }
      } else {
        if (xamr[1].first <= rl[i + j] + lr[n + i - j - 1] - arr[i][j]) {
          xamr[1].first = rl[i + j] + lr[n + i - j - 1] - arr[i][j];
          xamr[1].second = make_pair(i + 1, j + 1);
          if (xamr[0].first <= xamr[1].first) swap(xamr[0], xamr[1]);
        }
      }
    }
  }
  printf("%lld\n", xamr[0].first + xaml[0].first);
  printf("%d %d %d %d\n", xamr[0].second.first, xamr[0].second.second,
         xaml[0].second.first, xaml[0].second.second);
}
