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>
const int N = 2010;
using namespace std;
map<int, long long> mn, sc;
long long grid[N][N], res[2];
int x[2], y[2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      scanf("%lld", &grid[i][j]);
      mn[i + j] += grid[i][j];
      sc[i - j] += grid[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      int parity = (i + j) % 2;
      if (res[parity] <= (mn[i + j] + sc[i - j]) - grid[i][j]) {
        res[parity] = (mn[i + j] + sc[i - j]) - grid[i][j];
        x[parity] = i;
        y[parity] = j;
      }
    }
  }
  printf("%lld\n%d %d %d %d\n", res[0] + res[1], x[0], y[0], x[1], y[1]);
  return 0;
}
