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.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class P_463C {
    static final FS sc = new FS();
    static final PrintWriter pw = new PrintWriter(System.out);

    public static void main(String[] args) {
        int n = sc.nextInt();
        long[][] a = new long[n+1][n+1];
        long[][] m = new long[n+1][n+1];
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                a[i][j] = sc.nextLong();
            }
        }
        long[] sm = new long[2*n-1];
        long[] ss = new long[2*n-1];
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                sm[i+j-2] += a[i][j];
                ss[i-j+n-1] += a[i][j];
            }
        }
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                m[i][j] = sm[i+j-2] + ss[i-j+n-1] - a[i][j];
            }
        }
        long x1 = 0, y1 = 0, x2 = 0, y2 = 0;
        long odd = -1, even = -1;
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                if(((i+j)&1)==1){
                    if(odd<m[i][j]){
                        odd = m[i][j];
                        x1 = i; y1 = j;
                    }
                }
                else{
                    if(even<m[i][j]){
                        even = m[i][j];
                        x2 = i; y2 = j;
                    }
                }
            }
        }
        System.out.println(odd+even);
        System.out.println(x1+" "+y1+" "+x2+" "+y2);
    }

    static class FS {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception ignored) {
                }
            }
            return st.nextToken();
        }

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

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
            }
            return a;
        }

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

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