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.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;

public class quedt {

	public static void main(String[] args) throws IOException {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		long[][] arr = new long[n][n];
		HashMap<Integer, Long> left = new HashMap<>();
		HashMap<Integer, Long> right = new HashMap<>();
		
		for (int i = 0; i < n; i++) {
			String []str=br.readLine().split(" ");
			for (int j = 0; j < n; j++) {
				arr[i][j] = Long.parseLong(str[j]);
			}
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				long val=arr[i][j];
	     long pehle= left.getOrDefault(i-j, 0L);
	     left.put((i-j), pehle+val);
	     long pehle2= right.getOrDefault(n-1-(i+j), 0L);
	     right.put(n-1-(i+j), pehle2+val);
			}
		}
				long maxe=0,maxo=0;
		int x1=1,x2=1,y1=1,y2=2;
		int [][]dir={{0,-1},{-1,0},{0,1},{1,0}};
		 for(int i=0;i<n;i++){
			 for(int j=0;j<n;j++){
				long val= left.get((i-j))+right.get((n-1-(i+j)))-arr[i][j];
			 
			    if((i+j)%2==0){
			    	if(val>maxe){
			    		x1=i+1;
			    		y1=j+1;
			    		maxe=val;
			    	}
			    }else{
			    	if(val>maxo){
			    		x2=i+1;
			    		y2=j+1;
			    		maxo=val;
			    	}
			    }
			 }
		 }
		 System.out.println(maxe+maxo);
		 System.out.println(x1+" "+y1+" "+x2+" "+y2);
	}

}
