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.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
    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; 
        } 
    } 
    static int MOD=1000000000+7;
    
    //Brian Kernighan’s Algorithm
    static long countSetBits(long n){
        if(n==0) return 0;
        return 1+countSetBits(n&(n-1));
    }
    
    static long factorial(long n){
        if(n==1) return 1;
        if(n==2) return 2;
        if(n==3) return 6;
        return n*factorial(n-1);
    }
    //Euclidean Algorithm
    static long gcd(long A,long B){
        if(B==0) return A;
        return gcd(B,A%B);
    }
    
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2);
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2));
    }
    //AKS Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i*i<=n;i+=6)
            if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
    public static void main (String[] args) throws java.lang.Exception
    {
        FastReader sc=new FastReader();
        int n=sc.nextInt();
        int a[][]=new int[n][n];
        for(int i=0;i<n;i++)
        	for(int j=0;j<n;j++)a[i][j]=sc.nextInt();
        long p[]=new long[2*n-1],s[]=new long[2*n-1];
        Arrays.fill(p,0);
        Arrays.fill(s,0);
        int x=0;
        for(int i=0;i<n;i++){
        	for(int j=0;j<n;j++){
        		p[i+j]+=a[i][j];
        		s[i-j+n-1]+=a[i][j];
        	}
        }
        long max1=-1,max2=-1,p1X=-1,p1Y=-1,p2X=-1,p2Y=-1;
        int b=1;
        for(int i=0;i<n;i++){
        	b=b^1;
        	for(int j=b;j<n;j+=2){
		      	if((p[i+j]+s[i-j+n-1]-a[i][j])>max1) {
		      		max1=p[i+j]+s[i-j+n-1]-a[i][j];
		      		p1X=i+1;
		      		p1Y=j+1;
		      	}
        	}
        }
        b=0;
        for(int i=0;i<n;i++){
        	b=b^1;
        	for(int j=b;j<n;j+=2){
		      	if((p[i+j]+s[i-j+n-1]-a[i][j])>max2) {
		      		max2=p[i+j]+s[i-j+n-1]-a[i][j];
		      		p2X=i+1;
		      		p2Y=j+1;
		      	}
        	}
        }
        System.out.println(max1+max2);
        System.out.println(p1X+" "+p1Y+" "+p2X+" "+p2Y);
        
    }
}