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 p54 {
    static class FastReader{ 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                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()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }

    public static void main(String[] args){
        FastReader sc = new FastReader();
        int n = sc.nextInt();
        
        Map<Integer, Long> row = new HashMap<>();
        Map<Integer, Long> col = new HashMap<>();
        long[][] arr = new long[n+1][n+1];
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++){
                arr[i][j] = sc.nextLong();
                row.put(i+j, arr[i][j]+row.getOrDefault(i+j, 0L));
                col.put(i-j, arr[i][j]+col.getOrDefault(i-j, 0L));
            }

        long max1=-1, max2=-1;
        int r1=-1, c1=-1, r2=-1, c2=-1;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++){
                long val = row.getOrDefault(i+j, 0L) + col.getOrDefault(i-j, 0L) - arr[i][j];
                if(i%2==j%2){
                    if(val>max1){
                        max1 = val;
                        r1 = i;
                        c1 = j;
                        // System.out.println(r1+" "+c1+" "+max1);
                    }
                }else{
                    if(val>max2){
                        max2 = val;
                        r2 = i;
                        c2 = j;
                        // System.out.println(r2+" "+c2+" "+max2);
                    }
                }
            }

        System.out.println(max1+max2);
        System.out.println(r1+" "+c1+" "+r2+" "+c2);
    }
}