Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

4
west
east
wait

Sample Output 1

3

In this sample, initially A = west、B = east、C = wait. We can achieve the objective in the minimum number of operations by performing three operations as follows:

Change the second character in A to a. A is now wast.

Change the first character in B to w. B is now wast.

Change the third character in C to s. C is now wast.

Sample Input 2

9
different
different
different

Sample Output 2

0

If A, B and C are already equal in the beginning, the number of operations required is 0.

Sample Input 3

7
zenkoku
touitsu
program

Sample Output 3

13

Solution in C (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(void){
    int n,i,cnt;
    char a[100],b[100],c[100];
    scanf("%d",&n);
    scanf("%s",a);
    scanf("%s",b);
    scanf("%s",c);
    cnt=0;
    for(i=0;i<=n-1;i++){
        if(a[i]!=b[i]){
            if(a[i]!=c[i]){
                if(b[i]!=c[i]){
                    cnt=cnt+2;
                }
                else{
                    cnt=cnt+1;
                }
            }
            else{
                cnt=cnt+1;
            }
        }
        else{
            if(a[i]!=c[i]){
                cnt=cnt+1;
            }
        }
    }
    printf("%d",cnt);
}