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 n;
long long a[2002][2002];
long long d1[4010], d2[4010];
long long ans[2];
int main() {
  scanf("%d", &n);
  for (int register i = 0; i < n; i++)
    for (int register j = 0; j < n; j++) {
      scanf("%d", &a[i][j]);
      d1[i + j] += a[i][j];
      d2[i - j + n] += a[i][j];
    }
  int x1, y1, x2, y2;
  for (int register i = 0; i < n; i++)
    for (int register j = 0; j < n; j++) {
      ans[(i + j & 1)] =
          max(ans[(i + j & 1)], d1[i + j] + d2[i - j + n] - a[i][j]);
      if (ans[(i + j & 1)] == d1[i + j] + d2[i - j + n] - a[i][j]) {
        if ((i + j & 1))
          x1 = i, y1 = j;
        else
          x2 = i, y2 = j;
      }
    }
  printf("%I64d", ans[0] + ans[1]);
  printf("\n");
  printf("%d %d %d %d", x1 + 1, y1 + 1, x2 + 1, y2 + 1);
  return 0;
}
