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: Wrong Answer)

#include <iostream>
#include <string>
using namespace std;

int main () {

  int l, count = 0;
  string a, b, c;
  cin >> l >> a >> b >> c;

  for (int i = 0; i < l; i++) {
    if (a[i] == b[i] && a[i] == c[i] && b[i] == c[i]) {
      continue;
    }

    if (b.find(a[i]) != i) {
      if (c.find(a[i]) == i) {
        b[i] = a[i];
        count++;
        continue;
      }

      if (b.find(c[i]) == i) {
        a[i] = b[i];
        count++;
        continue;
      }
      b[i] = a[i];
      c[i] = a[i];
      count+=2;
      continue;
    } 
      c[i] = a[i];
      count++;

  }
  cout << count << endl;

  return 0;
}