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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class C {

    static StringTokenizer st;
    static BufferedReader br;
    static PrintWriter pw;
    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int n = nextInt();
        long[]d1 = new long[2*n], d2 = new long[2*n];
        int[][]a = new int[n+1][n+1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                a[i][j] = nextInt();
                d1[i+j-1] += a[i][j];
                d2[n-j+i] += a[i][j];
            }
        }
        long max1 = -1, max2 = -1;
        int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                long cur = d1[i+j-1]+d2[n-j+i]-a[i][j];
                if ((i+j) % 2==0) {
                    if (cur > max1) {
                        max1 = cur;
                        x1 = i;
                        y1 = j;
                    }
                }
                else {
                    if (cur > max2) {
                        max2 = cur;
                        x2 = i;
                        y2 = j;
                    }
                }
            }
        }
        System.out.println(max1+max2);
        System.out.println(x1+" "+y1+" "+x2+" "+y2);
        pw.close();
    }
    private static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    private static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
    private static String next() throws IOException {
        while (st==null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }
}
