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 (Clang 3.8.0) (judge verdict: Accepted)

#include<iostream>
#include<iomanip>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
const int inf=2147483647,dx[]={-1,0,1,0},dy[]={0,-1,0,1};// 上 左 下 右
const int N=100005,M=1000005,mod=1000000007;
const long long llinf=9223372036854775807ll;
int n,cnt[105][30],maxv,ans;
string a,b,c;
int main(){
	cin>>n>>a>>b>>c;
	for(int i=0;i<n;i++)
	{
		maxv=0;
		cnt[i][a[i]-'a']++,cnt[i][b[i]-'a']++,cnt[i][c[i]-'a']++;
		for(int j=0;j<26;j++)
			maxv=max(maxv,cnt[i][j]);
		ans+=(3-maxv);
	}
	cout<<ans;
	return 0;
}