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>
long long n, i, j, a, arr[3000][3000], G[2005], H[10005], max1, max2,
    max1x = 1, max1y = 1, max2x = 1, max2y = 2, temp;
int main() {
  scanf("%lld", &n);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      scanf("%lld", &a);
      H[i - j + 3000] += a;
      G[i + j] += a;
      arr[i][j] = a;
    }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      temp = H[i - j + 3000] + G[i + j] - arr[i][j];
      if ((i + j) % 2 == 0) {
        if (temp > max1) max1 = temp, max1x = i, max1y = j;
      } else {
        if (temp > max2) max2 = temp, max2x = i, max2y = j;
      }
    }
  printf("%lld\n%lld %lld %lld %lld", max1 + max2, max1x, max1y, max2x, max2y);
  return 0;
}
