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>
struct rec {
  int first;
  int second;
} v[2];
using namespace std;
int a[2002][2002];
long long d1[5004], d2[5004], sol[2];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
      d1[i + j] += a[i][j];
      d2[i - j + n] += a[i][j];
    }
  }
  sol[0] = -1;
  sol[1] = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      int c = (i + j) & 1;
      long long val = d1[i + j] + d2[i - j + n] - a[i][j];
      if (val > sol[c]) {
        sol[c] = val;
        v[c].first = i;
        v[c].second = j;
      }
    }
  }
  cout << sol[0] + sol[1] << endl;
  cout << v[0].first << " " << v[0].second << " ";
  cout << v[1].first << " " << v[1].second << "\n";
  return 0;
}
