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;
unsigned long long a[2002][2002], d1[4003], d2[4003], i, j, k, n;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  memset(d1, 0, sizeof(d1));
  memset(d2, 0, sizeof(d2));
  unsigned long long m = 0, x, y, x1, y1, m2 = 0;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      cin >> a[i][j];
      d2[i + j] += a[i][j];
      if (i >= j) {
        d1[i - j] += a[i][j];
      } else {
        d1[2 * n - 1 + i - j] += a[i][j];
      }
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (i >= j) {
        a[i][j] = d2[i + j] + d1[i - j] - a[i][j];
      } else {
        a[i][j] = d2[i + j] + d1[2 * n - 1 + i - j] - a[i][j];
      }
    }
  }
  m = m2 = 0;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        if (m != max(m, a[i][j])) {
          m = max(m, a[i][j]);
          x = i;
          y = j;
        }
      } else if (m2 != max(m2, a[i][j])) {
        m2 = max(m2, a[i][j]);
        x1 = (unsigned long long)i;
        y1 = (unsigned long long)j;
      }
    }
  }
  if (m + m2 == 0) {
    cout << "0" << endl;
    cout << "1"
         << " "
         << "1"
         << " "
         << "2"
         << " "
         << "1";
  } else {
    cout << m + m2 << endl;
    cout << x << " " << y << " " << x1 << " " << y1 << endl;
  }
  return 0;
}
