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;
const long long maxsize = 10000 + 9;
long long primary[maxsize], secondary[maxsize], sum[2010][2010], a[2010][2010];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  T = 1;
  while (T--) {
    long long i, j, k, l, m1 = 0, m2 = 0, n, x1 = 1, x2 = 1, y1 = 1, y2 = 2;
    cin >> n;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        cin >> a[i][j];
        primary[i - j + n] += a[i][j];
        secondary[i + j] += a[i][j];
        sum[i][j] += primary[i - j + n] + secondary[i + j] - a[i][j];
      }
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        sum[i][j] = primary[i - j + n] + secondary[i + j] - a[i][j];
      }
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if ((i + j) & 1) {
          if (m1 < sum[i][j]) {
            m1 = sum[i][j];
            x1 = i;
            y1 = j;
          }
        } else {
          if (m2 < sum[i][j]) {
            m2 = sum[i][j];
            x2 = i;
            y2 = j;
          }
        }
      }
    }
    cout << m1 + m2 << '\n';
    cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1 << " "
         << '\n';
  }
}
