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 <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 105
#define ll long long
#define ld double
#define mod 998244353
using namespace std;
char inp[3][maxn];
int main()
{
	int n;
	scanf("%d", &n);
	for(int i = 0; i < 3; i++)
		scanf("%s", inp[i]);
	int res = 0;
	for(int j = 0; j < n; j++)
	{
		int ns[26];
		memset(ns, 0, sizeof(ns));
		int ans = 0;
		for(int l = 0; l < 3; l++)
			if(!ns[inp[l][j] - 'a'])
				ans++, ns[inp[l][j] - 'a'] = 1;
		res += ans - 1;
	}
	printf("%d\n", res);
	return 0;
}