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>

int main()
{
    int n,i,j;
    char abc[3][101];

    scanf("%d",&n);
    for (i = 0; i < 3; i++)
    {
        scanf("%s",abc[i]);
    }
    
    int buf = 0,ans = 0;
    
    for (i = 0; i < n; i++)
    {
        if(abc[0][i] != abc[1][i] && abc[1][i] != abc[2][i] && abc[2][i] != abc[0][i])
        {
            ans += 2;
        }
        else if(abc[0][i] == abc[1][i] && abc[1][i] == abc[2][i] && abc[2][i] == abc[0][i])
        {
            ans += 0;
        }
        else
        {
            ans += 1;
        }
        
    }
    

    printf("%d",ans);
}