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>
int main() {
  int n, **p, i, j;
  long long int t, max1 = 0, max2 = 0, s1[5000] = {0}, s2[5000] = {0};
  scanf("%d", &n);
  p = (int **)malloc(n * sizeof(int *));
  for (i = 0; i < n; i++) p[i] = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      scanf("%d", &p[i][j]);
      s1[i + j] += p[i][j];
      s2[i - j + n] += p[i][j];
    }
  int x1 = 0, y1 = 0, x2 = 0, y2 = 1;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      t = s1[i + j] + s2[i - j + n] - p[i][j];
      if ((i + j) % 2 == 0 && t > max1) {
        x1 = i;
        y1 = j;
        max1 = t;
      }
      if ((i + j) % 2 == 1 && t > max2) {
        x2 = i;
        y2 = j;
        max2 = t;
      }
    }
  printf("%I64d\n", max1 + max2);
  printf("%d %d %d %d\n", x1 + 1, y1 + 1, x2 + 1, y2 + 1);
  return 0;
}
