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
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class C {

  public static void main(String[] args) throws NumberFormatException, IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    int n = Integer.parseInt(br.readLine());
    int[][] tb = new int[n][n];

    for (int i = 0; i < n; i++) {
      StringTokenizer tokenizer = new StringTokenizer(br.readLine());
      for (int j = 0; j < n; j++) {
        tb[i][j] = Integer.parseInt(tokenizer.nextToken());
      }
    }

    long[] di1 = new long[n * 2 - 1];
    long[] di2 = new long[n * 2 - 1];

    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        di1[i + j] += tb[i][j];
        di2[i - j + n - 1] += tb[i][j];
      }
    }

    long[] bscores = new long[] { -1, -1 };
    int[] bis = new int[2];
    int[] bjs = new int[2];

    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        int k = (i + j) % 2;
        long score = di1[i + j] + di2[i - j + n - 1] - tb[i][j];
        if (score > bscores[k]) {
          bscores[k] = score;
          bis[k] = i + 1;
          bjs[k] = j + 1;
        }
      }
    }

    System.out.printf("%d\n%d %d %d %d", bscores[0] + bscores[1],
        bis[0], bjs[0], bis[1], bjs[1]);
  }

}