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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>
#include <stdlib.h>
int main(){
	int n=0;
	char a[110];
	char b[110];
	char c[110];
	scanf("%d\n",&n);
	for(int i=0;i<110;i++){
		a[i]='$';
		b[i]='$';
		c[i]='$';
	}
		scanf("%s\n",a);
		scanf("%s\n",b);
		scanf("%s\n",c);

/*	for(int i=0;i<n-1;i++){
		scanf("%s ",a[i]);
	}
		scanf("%s\n",a[n-1]);
	for(int i=0;i<n-1;i++){
		scanf("%s ",b[i]);
	}
	scanf("%s\n",b[n-1]);
	for(int i=0;i<n-1;i++){
		scanf("%s ",c[i]);
	}
		scanf("%s\n",c[n-1]);
*/
	int ans=0;
	for(int i=0;i<n;i++){
		ans=ans+2;
		if(a[i]==b[i]||b[i]==c[i]||c[i]==a[i]){
			ans--;
		}
		if(a[i]==b[i]&&b[i]==c[i]&&c[i]==a[i]){
			ans--;
		}
	}
	printf("%d\n",ans);

}