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 <algorithm>
#include <cstdio>
#include <deque>
#include <iomanip>
#include <iostream>
#include <queue>
#include <stack>
#include <utility>
#include <vector>
using std::cin;
using std::cout;
using std::endl;

int main(void) {
  cout << std::fixed << std::setprecision(10);
  cin.tie(0);
  std::ios::sync_with_stdio(false);
  int n;
  std::string a, b, c;
  cin >> n >> a >> b >> c;
  int res = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] && b[i] == c[i]) {
      // nothing
    } else if (a[i] == b[i] || b[i] == c[i] || a[i] == c[i]) {
      res++;
    } else {
      res += 2;
    }
  }
  cout << res << endl;

  return 0;
}