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.io.PrintWriter;


public class CF_264_C {
	public static void main(String[] args) throws IOException{
		PrintWriter pw = new PrintWriter(System.out, true);
		BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(input.readLine());
		int[][] map = new int[n][n];
		long[][] res = new long[n][n];
		String[] s;
		for(int i=0;i<n;i++){
			s = input.readLine().split(" ");
			for(int j=0;j<n;j++){
				map[i][j] = Integer.parseInt(s[j]);
			}
		}
		long[] a = new long[2*n], b = new long[2*n];
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				a[j+i] += map[i][j];
				b[j-i+n-1] += map[i][j];
			}
		}
		
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				res[i][j] += a[i+j]+b[j-i+n-1]-map[i][j];
			}
		}
		long max = Long.MIN_VALUE, max1 = Long.MIN_VALUE;
		int start = 1, x1 = 0, y1 = 0, x2 = 0, y2 = 0;
		
		for(int i=0;i<n;i++){
			start ^= 1;
			for(int j=start;j<n;j+=2){
				if(max < res[i][j]){
					max = res[i][j];
					x1 = i;
					y1 = j;
				}
			}
		}
		start = 0;
		for(int i=0;i<n;i++){
			start ^= 1;
			for(int j=start;j<n;j+=2){
				if(max1 < res[i][j]){
					max1 = res[i][j];
					x2 = i;
					y2 = j;
				}
			}
		}
		x1++;
		y1++;
		x2++;
		y2++;
			
		pw.println((max+max1));
		pw.println(x1+" "+y1+" "+x2+" "+y2);
		pw.close();
		input.close();
	}
}
