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 {
    static boolean DEBUG_FLAG = false;
    int INF = (int)1e9;
    long MOD = 1000000007;
    long M = (long)1e18;
    
    static void debug(String s) {
        if(DEBUG_FLAG) {
            System.out.print(s);
        }
    }
    
    void solve(InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int[][] a = new int[n+2][n+2];
        long[] d1 = new long[2*n+10];
        long[] d2 = new long[2*n+10];
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=n; j++) {
                a[i][j] = in.nextInt();
                d1[i+j] += a[i][j];
                d2[i-j+n] += a[i][j];
            }
        }
        int x1=-1, y1=-1, x2=-1, y2=-1;
        long e = -1, o = -1;
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=n; j++) {
                long v = d1[i+j] + d2[i-j+n] - a[i][j];
                if((i+j)%2==0) {
                    if(e < v) {
                        e = v;
                        x1 = i;
                        y1 = j;
                    }
                } else {
                    if(o < v) {
                        o = v;
                        x2 = i;
                        y2 = j;
                    }
                }
            }
        }
        out.println(e+o);
        out.println(x1+" "+y1+" "+x2+" "+y2);
    }
    
    
    public static void main(String[] args) throws IOException {
        if(args.length>0 && args[0].equalsIgnoreCase("d")) {
            DEBUG_FLAG = true;
        }
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt();
        long start = System.nanoTime();
        while(t-- >0) {
            new test().solve(in, out);
        }
        long end = System.nanoTime();
        debug("\nTime: " + (end-start)/1e6 + " \n\n");
        out.close();
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
    
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}