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>
#define MOD 1000000007
typedef long long int ll;
void llswap(ll *x,ll *y){ll tmp;tmp=*x;*x=*y;*y=tmp;}
void swap(int *x,int *y){int tmp;tmp=*x;*x=*y;*y=tmp;}
int rmax(int x,int y){return x>y?x:y;}
int rmin(int x,int y){return x>y?y:x;}
ll llrmax(ll x,ll y){return x>y?x:y;}
ll llrmin(ll x,ll y){return x>y?y:x;}
int asc(const void *a,const void *b){return *(int*)a-*(int*)b;}
int desc(const void *a,const void *b){return *(int*)b-*(int*)a;}

int main(){
    int n;
    char abc[3][200];

    scanf("%d",&n);
    for(int i=0;i<3;i++){
        scanf("%s",abc[i]);
    }

    int dif[200][26]={{}};

    int ans=0,flg=0;
    int cnt=0;
    for(int i=0;i<n;i++){
        flg=0;
        cnt=0;
        for(int j=0;j<2;j++){
            for(int k=j+1;k<3;k++){
                if(abc[j][i]==abc[k][i]){
                    cnt++;
                }
            }
        }

        //printf("cnt:%d\n",cnt);
        if(cnt>1){
            
        }else if(cnt==1){
            ans++;
        }else{
            ans+=2;
        }
    }

    printf("%d\n",ans);
    return 0;
}