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 Test {

    private static int c = 1000_000_007;

    public static void main(String[] args) {
//        Scanner in = new Scanner(System.in);
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int n;
        n = in.nextInt();
        int[][] a = new int[n][n];
        long[] a1, a2, b1, b2;
        a1 = new long[2 * n];
        a2 = new long[2 * n];
        b1 = new long[2 * n];
        b2 = new long[2 * n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                a[i][j] = in.nextInt();
                if ((i + j) % 2 == 0) {
                    a1[i + j] += a[i][j];
                    a2[i + n - 1 - j] += a[i][j];
                } else {
                    b1[i + j] += a[i][j];
                    b2[i + n - 1 - j] += a[i][j];
                }
            }
        }
        int x1 = 1, y1 = 1, x2 = 1, y2 = 2;
        long maxa = 0, maxb = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if ((i + j) % 2 == 0) {
                    if ((a1[i + j]+a2[i + n - 1 - j]-a[i][j]) > maxa) {
                        maxa = a1[i + j]+a2[i + n - 1 - j]-a[i][j];
                        x1 = i+1; y1 = j+1;
                    }
                    
                } else {
                    if ((b1[i + j]+b2[i + n - 1 - j]-a[i][j]) > maxb) {
                        maxb = b1[i + j]+b2[i + n - 1 - j]-a[i][j];
                        x2 = i+1; y2 = j+1;
                    }
                }
            }
        }
        System.out.println(maxa+maxb);
        System.out.print(x1+" ");
        System.out.print(y1+" ");
        System.out.print(x2+" ");
        System.out.println(y2);
    }

}

class FastScanner {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String next() {
        while (!st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    int[] readArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    long nextLong() {
        return Long.parseLong(next());
    }
}
