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, c[2010][2010], x[2], y[2];
long long a[4010], b[4010], ret[2];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      scanf("%d", &c[i][j]);
      a[i + j] += c[i][j];
      b[i - j + n] += c[i][j];
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (a[i + j] + b[i - j + n] - c[i][j] >= ret[(i + j) & 1]) {
        int v = (i + j) & 1;
        ret[v] = a[i + j] + b[i - j + n] - c[i][j];
        x[v] = i + 1;
        y[v] = j + 1;
      }
  printf("%I64d\n", ret[0] + ret[1]);
  printf("%d %d %d %d\n", x[0], y[0], x[1], y[1]);
}
