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;
long long int inf = 1e9;
const long long int mod = 1e9 + 7;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<vector<long long int> > a(n, vector<long long int>(n));
  vector<long long int> d1(2 * n - 1, 0), d2(2 * n - 1, 0);
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      cin >> a[i][j];
      d1[i + j] += a[i][j];
      d2[i - j + n - 1] += a[i][j];
    }
  }
  long long int max1 = 0, max2 = 0;
  pair<long long int, long long int> ans1, ans2;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      if ((i + j) % 2 == 0) {
        if (max1 <= d1[i + j] + d2[i - j + n - 1] - a[i][j]) {
          max1 = d1[i + j] + d2[i - j + n - 1] - a[i][j];
          ans1 = make_pair(i + 1, j + 1);
        }
      } else {
        if (max2 <= d1[i + j] + d2[i - j + n - 1] - a[i][j]) {
          max2 = d1[i + j] + d2[i - j + n - 1] - a[i][j];
          ans2 = make_pair(i + 1, j + 1);
        }
      }
    }
  }
  cout << max1 + max2 << '\n';
  cout << ans1.first << " " << ans1.second << " " << ans2.first << " "
       << ans2.second << '\n';
}
