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<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr int Inf = 1000000001;
constexpr ll INF= 1e18;
constexpr ll MOD = 1000000007;
const double PI = 3.1415926535897;
typedef pair<int,int> P;

int main() {
    int N;
    cin >> N;
    string A,B,C;
    cin >> A >> B >> C;
    int ret = 0;
    for(int i = 0;i < N;i++) {
        if(A.at(i) == B.at(i) && B.at(i) == C.at(i)) {
            ret = ret;
        }
        else if(A.at(i) == B.at(i)) {
            ret++;
        }
        else if(B.at(i) == C.at(i)) {
            ret++;
        }
        else if(A.at(i) == C.at(i)) {
            ret++;
        }
        else {
            ret += 2;
        }
    }
    cout << ret << endl;
}