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.*;
import java.lang.*;
import java.math.BigInteger;


public class Main {
  
    public static long[][] a;
    public static  int n;
	public static void main(String[] args) throws Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t,ca;
		String[] my;
		String line = br.readLine();
	    int n = Integer.parseInt(line);
	    long d1[] = new long[2*n+1];
	    long d2[] = new long[2*n+1];
	    a = new long[n+1][n+1];
	    int i,j;
	    for(i=1;i<=n;i++)
	    {
	        line = br.readLine();
	        my = line.split(" ");
	        for(j=1;j<=n;j++)
	        {
	            a[i][j] = Long.parseLong(my[j-1]);
	            d1[i+j] +=a[i][j];
	            d2[i-j+n] += a[i][j];
	        }
	        
	    }
	    
	    long fir,sec;
	    fir = sec = -1;
	    int x1,y1,x2,y2;
	    x1 = y1 = x2 = y2 = -1;
	    for(i=1;i<=n;i++)
	    {
	        for(j=1;j<=n;j++)
	        {
	            long cur = d1[i+j] + d2[i-j+n] - a[i][j];
	            int x = i+j;
	            if(x%2==0)
	            {
	                if(cur > fir){
	                    fir = cur;
	                    x1 = i;
	                    y1 = j;
	                }
	            }
	            else {
	                if(cur>sec)
	                {
	                    sec = cur;
	                    x2 = i;
	                    y2 = j;
	                }
	            }
	        }
	    }
	    System.out.println(fir + sec);
	    System.out.println(x1 + " " + y1 + " " + x2 + " " + y2);

        }
    
}

