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;
const int MOD(1000000007);
const int INF((1 << 30) - 1);
const int MAXN(2005);
int board[MAXN][MAXN];
long long a[2 * MAXN], b[2 * MAXN];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      int x;
      scanf("%d", &x);
      board[i][j] = x;
      a[i + j] += x;
      b[i - j + n - 1] += x;
    }
  long long ans[2] = {-1, -1};
  int x[2], y[2];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      int c = (i + j) & 1;
      if (a[i + j] + b[i - j + n - 1] - board[i][j] > ans[c]) {
        ans[c] = a[i + j] + b[i - j + n - 1] - board[i][j];
        x[c] = i;
        y[c] = j;
      }
    }
  printf("%I64d\n", ans[0] + ans[1]);
  printf("%d %d %d %d", x[0] + 1, y[0] + 1, x[1] + 1, y[1] + 1);
}
