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 arr[2001][2001];
long long d1[4050], d2[4050];
int n;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      int x;
      cin >> arr[i][j];
      d1[i + j] += arr[i][j];
      d2[n + i - j] += arr[i][j];
    }
  }
  long long maxi = -1;
  long long maxi2 = -1;
  int pos1, pos2, pos3, pos4;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i + j) % 2) {
        if (d1[i + j] + d2[n + i - j] - arr[i][j] > maxi)
          maxi = d1[i + j] + d2[n + i - j] - arr[i][j], pos1 = i + 1,
          pos2 = j + 1;
      } else if ((i + j) % 2 == 0) {
        if (d1[i + j] + d2[n + i - j] - arr[i][j] > maxi2)
          maxi2 = d1[i + j] + d2[n + i - j] - arr[i][j], pos3 = i + 1,
          pos4 = j + 1;
      }
    }
  }
  cout << maxi + maxi2 << endl
       << pos1 << " " << pos2 << " " << pos3 << " " << pos4 << endl;
  return 0;
}
