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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
int main(void){
    int length;
    char *A,*B,*C;
    scanf("%d",&length);
    A=(char *)malloc(sizeof(char)*length);
    B=(char *)malloc(sizeof(char)*length);
    C=(char *)malloc(sizeof(char)*length);
    scanf("%s",A);
    scanf("%s",B);
    scanf("%s",C);
    int count1=0,count2=0;
    int i;
    for(i=0;i<length;i++){
        if(A[i]==B[i]&&B[i]==C[i]){
            count1++;
        }
        else if(A[i]==B[i]&&B[i]!=C[i]){
            count2++;
        }
        else if(A[i]==C[i]&&B[i]!=C[i]){
            count2++;
        }
        else if(C[i]==B[i]&&A[i]!=C[i]){
            count2++;
        }
    }
    //printf("%d %d\n",count1,count2);
    printf("%d",count2+2*(length-count1-count2));
    return 0;
}