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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class Main {

	public static void main(String[] args) throws IOException 
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt(), a[][] = new int[n][n];
		long[] d1 = new long[2*n-1], d2 = new long[2*n-1];
		for(int i = 0; i < n; ++i)
			for(int j = 0; j < n; ++j)
			{
				a[i][j] = sc.nextInt();
				d1[i+j] += a[i][j];
				d2[i-j+n-1] += a[i][j];
			}

		int x1 = -1, x2 = -1, y1 = -1, y2 = -1;
		long max1 = -1, max2 = -1;
		for(int i = 0; i < n; ++i)
			for(int j = 0; j < n; ++j)
				{
					long cur = d1[i+j] + d2[i-j+n-1]- a[i][j];
					if((i+j)%2 == 0)
					{
						if(cur > max1)
						{
							max1 = cur;
							x1 = i;
							y1 = j;
						}
					}
					else
					{
						if(cur > max2)
						{
							max2 = cur;
							x2 = i;
							y2 = j;
						}
					}
				}
		out.printf("%d\n%d %d %d %d\n", max1 + max2, x1 + 1, y1 + 1, x2 + 1, y2 + 1);
		out.flush();
		out.close();

	}



	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {return Integer.parseInt(next());}

		public long nextLong() throws IOException {return Long.parseLong(next());}

		public String nextLine() throws IOException {return br.readLine();}

		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}

		public boolean ready() throws IOException {return br.ready();}


	}
}