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.*;

public class C1
{
	public static void main(String[] args) throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = Integer.parseInt(br.readLine());
		long[][] grid = new long[n][n];
		for(int i = 0; i < n; i++)
		{
			StringTokenizer st = new StringTokenizer(br.readLine());
			for(int j = 0; j < n; j++)
				grid[i][j] = Long.parseLong(st.nextToken());
		}
		long[] left = new long[2*n-1], right = new long[2*n-1];
		for(int i = 0; i < n; i++)
			for(int j = 0; j < n; j++)
			{
				left[i+j] += grid[i][j];
				right[i+n-1-j] += grid[i][j];
			}
		long[][] bSum = new long[n][n];
		for(int i = 0; i < n; i++)
			for(int j = 0; j < n; j++)
				bSum[i][j] = left[i+j] + right[i+n-1-j] - grid[i][j];
		long max1 = Long.MIN_VALUE;
		int x1 = -1, y1 = -1;
		for(int i = 0; i < n; i++)
			for(int j = i%2; j < n; j += 2)
				if(bSum[i][j] > max1)
				{
					max1 = bSum[i][j];
					x1 = i;
					y1 = j;
				}
		long max2 = Long.MIN_VALUE;
		int x2 = -1, y2 = -1;
		for(int i = 0; i < n; i++)
			for(int j = 1-i%2; j < n; j += 2)
				if(bSum[i][j] > max2)
				{
					max2 = bSum[i][j];
					x2 = i;
					y2 = j;
				}
		pw.println(max1 + max2 + "\n" + (x1+1) + " " + (y1+1) + " " + (x2+1) + " " + (y2+1));
		br.close(); pw.close(); System.exit(0);
	}
}