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 C {

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

		int n = sc.nextInt();

		long[][] mat = new long[n][n];

		long diagL[] = new long[n << 1], diagR[] = new long[n << 1];
		
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++) {
				mat[i][j] = sc.nextInt() ; 
				diagL[n - 1 + i - j] += mat[i][j] ;
				diagR[i + j] += mat[i][j];
			}
		
		long max1 = -1, max2 = -1, x1 = 0, y1 = 0, y2 = 0, x2 = 0;

		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++) {
				if (max1 <= diagL[n - 1 + i - j] + diagR[i + j] - mat[i][j]) {
					max1 = diagL[n - 1 + i - j] + diagR[i + j] - mat[i][j];
					x1 = i;
					y1 = j;
				}

			}

		for(int i = 0 ; i < n ;i++)
			for(int j = 0 ; j < n ; j++)
			{
				if(n - 1 + i - j == n - 1 + x1 - y1)
					diagR[i + j] = -1 ; 
				
				if(i + j == x1 + y1)
					diagL[n - 1 + i - j] = -1 ;
			}
		
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++) 
			{
				if (max2 <= diagL[n - 1 + i - j] + diagR[i + j] - mat[i][j] && diagL[n - 1 + i - j] != -1 && diagR[i + j] != -1) 
				{
					max2 = diagL[n - 1 + i - j] + diagR[i + j] - mat[i][j];
					x2 = i;
					y2 = j;
				}
			}
		
		x1++ ; y1++ ; x2++;y2++;
		
		System.out.println(max1 +max2);
		System.out.println(x1+" "+y1+" "+x2+" "+y2);

	}
	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

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

		public Scanner(String path) throws Exception {
			br = new BufferedReader(new FileReader(path));
		}

		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 double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

	}


}