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 main() {
  cin.sync_with_stdio(false);
  int n;
  cin >> n;
  vector<vector<long long>> A(n + 1, vector<long long>(n + 1));
  vector<long long> primary(2 * n, 0);
  vector<long long> secondary(2 * n, 0);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      cin >> A[i][j];
      primary[i - j + n] += A[i][j];
      secondary[i + j - 1] += A[i][j];
    }
  long long max1 = -9999, max2 = -99999, pos_i, pos_j, score = 0;
  long long pos2_i, pos2_j;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if ((i + j) % 2 &&
          primary[i - j + n] + secondary[i + j - 1] - A[i][j] > max1) {
        max1 = primary[i - j + n] + secondary[i + j - 1] - A[i][j];
        pos_i = i, pos_j = j;
      } else if ((i + j) % 2 == 0 &&
                 primary[i - j + n] + secondary[i + j - 1] - A[i][j] > max2) {
        max2 = primary[i - j + n] + secondary[i + j - 1] - A[i][j];
        pos2_i = i, pos2_j = j;
      }
    }
  score = max1 + max2;
  cout << score << endl;
  cout << pos_i << " " << pos_j << " " << pos2_i << " " << pos2_j << endl;
}
