You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters.

Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation:

* Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter.



What is the minimum number of operations required to achieve the objective?

Constraints

* 1 \leq N \leq 100
* Each of the strings A, B and C is a string of length N.
* Each character in each of the strings A, B and C is a lowercase English letter.

Input

Input is given from Standard Input in the following format:


N
A
B
C


Output

Print the minimum number of operations required.

Examples

Input

4
west
east
wait


Output

3


Input

9
different
different
different


Output

0


Input

7
zenkoku
touitsu
program


Output

13
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();

		String astr = sc.next();
		String bstr = sc.next();
		String cstr = sc.next();

		int count = 0;

		for(int i=0; i<n; i++) {
			if(astr.substring(i, i+1).equals(bstr.substring(i, i+1)) &&
					astr.substring(i, i+1).equals(cstr.substring(i, i+1))) {
				continue;
			} else if(astr.substring(i, i+1).equals(bstr.substring(i, i+1)) &&
					!(astr.substring(i, i+1).equals(cstr.substring(i, i+1)))) {
				count = count + 1;
			} else if(!(astr.substring(i, i+1).equals(bstr.substring(i, i+1))) &&
					astr.substring(i, i+1).equals(cstr.substring(i, i+1))) {
				count = count + 1;
			} else if(!(astr.substring(i, i+1).equals(bstr.substring(i, i+1))) &&
					!(astr.substring(i, i+1).equals(cstr.substring(i, i+1))) &&
					bstr.substring(i, i+1).equals(cstr.substring(i, i+1))) {
				count = count + 1;
			} else {
				count = count + 2;
			}
		}

		System.out.println(count);
		sc.close();
	}

}
