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 LIST
#include <cstdio>
#include <bitset>
#include <iostream>
#include <queue>
#include <stack>
#include <string>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <list>
#include <vector>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <iomanip>
using namespace std;

int main() {
  string a[10];
  int n;
  int ret = 0;
  
  cin >> n;
  for (int i = 0; i < 3; i++) cin >> a[i];
  
  for (int i = 0; i < n; i++) {
    int alph[100];
    memset(alph, 0, sizeof alph);
    
    for (int j = 0; j < 3; j++) {
      alph[a[j][i] - 'a']++;
    }
    
    int maks = 0;
    for (int j = 'a'; j <= 'z'; j++) {
      maks = max(maks, alph[j - 'a']);
    }
    
    ret += 3 - maks;
  }
  
  cout << ret << endl;
  return 0;
}