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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

int main(void){
   
   int N;
   char A[200],B[200],C[200];
   
   scanf("%d",&N);
   scanf("%s",A);
   scanf("%s",B);
   scanf("%s",C);
   
   int i;
   int count = 0;
   for(i = 0;i < N;i++){
      if(A[i] == B[i] && B[i] == C[i]){
         //Do nothing
      }
      else if(A[i] == B[i] && B[i] != C[i]){
         count = count + 1;
      }
      else if(A[i] != B[i] && B[i] == C[i]){
         count = count + 1;
      }
      else if(A[i] == C[i] && A[i] != B[i]){
         count = count + 1;
      }
      else if(A[i] != B[i] && B[i] != C[i] && A[i] != C[i]){
         count = count + 2;
      }
      else {
         printf("Error\n");
         exit(1);
      }
   }

   printf("%d\n",count);
   
   return 0;
}