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.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        // Scanner s = new Scanner(System.in);
        // int n = s.nextInt();
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());

        int a[][] = new int[n][n];
        long d1[] = new long[n * 2 - 1];
        long d2[] = new long[n * 2 - 1];

        for (int i = 0; i < n; i++) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            for (int j = 0; j < n; j++) {
                // a[i][j] = s.nextInt();
                a[i][j] = Integer.parseInt(st.nextToken());
                d1[i + j] += a[i][j];

                d2[n - i - 1 + j] += a[i][j];
            }
        }

        long m1 = 0, m2 = 0;
        int x1 = 0, y1 = 0;
        int x2 = 0, y2 = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                long r = d1[i + j] + d2[n - i - 1 + j] - a[i][j];
                if ((i + j) % 2 == 0) {
                    if (r > m1) {
                        m1 = r;
                        x1 = i;
                        y1 = j;
                    }
                } else {
                    if (r > m2) {
                        m2 = r;
                        x2 = i;
                        y2 = j;
                    }
                }
            }
        }

        System.out.println(m1 + m2);
        System.out.printf("%d %d %d %d", (x1 + 1), (y1 + 1), (x2 + 1), (y2 + 1));
    }
}