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>
#define rep(i,m) for(long long i=0; i<m; i++)
#define per(i,m) for(long long i=m-1; i>=0; i--)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define ROF(i,n,m) for(long long i=n-1; i>=m; i--)
#define all(x) (x).begin(),(x).end()
#define EPS (1e-7)
#define INF (1e18)
#define PI (acos(-1))
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
typedef pair<ll, ll> LP;

ll POW(ll x,ll n){
  if(n==0)return 1;
  if(n%2==0)return POW(x*x%MOD,n/2);
  return x*POW(x,n-1)%MOD;
}

ll gcd(ll u, ll v) {
  ll r;
  while (0 != v) {
    r = u % v; u = v; v = r;
  }
  return u;
}
 
ll KAI(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI(m-1)%MOD;
}
 
ll extGCD(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extGCD(b, a%b, y, x);
    y -= a / b * x;
    return d;
}
 
inline ll mod(ll a, ll m) {
    return (a % m + m) % m;
}
 
ll modinv(ll a) {
    ll x, y;
    extGCD(a, MOD, x, y);
    return mod(x, MOD);
}
 
ll COM(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI(m)%MOD*modinv(KAI(n)%MOD*KAI(m-n)%MOD)%MOD;
}

ll DEC(ll x,ll m,ll n)
{
  return x%POW(m,n+1)/POW(m,n);
}

ll keta(ll x,ll n)
{
  if(x==0)return 0;
  return keta(x/n,n)+1;
}

ll DIV(ll x,ll n)
{
  if(x==0)return 0;
  return x/n+DIV(x/n,n);
}


int main()
{
  ll n,ans;
  ans=0;
  char a[200],b[200],c[200];
  cin >> n;
  rep(i,n)cin >> a[i];
  rep(i,n)cin >> b[i];
  rep(i,n)cin >> c[i];
  rep(i,n){if(a[i]==b[i]&&b[i]==c[i])ans+=0;
           else if(a[i]==b[i]||b[i]==c[i]||a[i]==c[i])ans+=1;
           else ans+=2;
          }
  printf("%lld",ans);  
}