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() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<long long> V1(n * 2 - 1, 0), V2(n * 2 - 1, 0);
  vector<vector<long long>> V(n, vector<long long>(n));
  long long num;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      cin >> num;
      V1[n + i - j - 1] += num;
      V2[i + j] += num;
      V[i][j] = num;
    }
  long long ans1 = -1, ans2 = -1, x1, x2, y1, y2;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (((i & 1) && (j & 1)) || (!(i & 1) && !(j & 1))) {
        if (V1[n + i - j - 1] + V2[i + j] - V[i][j] > ans1)
          ans1 = V1[n + i - j - 1] + V2[i + j] - V[i][j], x1 = i, y1 = j;
      } else {
        if (V1[n + i - j - 1] + V2[i + j] - V[i][j] > ans2)
          ans2 = V1[n + i - j - 1] + V2[i + j] - V[i][j], x2 = i, y2 = j;
      }
    }
  cout << ans1 + ans2 << endl
       << x1 + 1 << ' ' << y1 + 1 << ' ' << x2 + 1 << ' ' << y2 + 1;
  return 0;
}
