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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author hakamada
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Nikkei_2 solver = new Nikkei_2();
        solver.solve(1, in, out);
        out.close();
    }

    static class Nikkei_2 {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int N = in.nextInt();
            String[] A = in.next().toUpperCase().split("");
            String[] B = in.next().toUpperCase().split("");
            String[] C = in.next().toUpperCase().split("");

            int count = 0;
            for (int i = 0; i < N; i++) {
                char cA = A[i].charAt(0);
                char cB = B[i].charAt(0);
                char cC = C[i].charAt(0);

                if ((cA == cB) && (cB == cC)) {
                } else if ((cA != cB) && (cB != cC) && (cA != cC)) {
                    count += 2;
                } else {
                    count++;
                }

            }
            out.println(count);
        }

    }
}

