Problem:
Score : 200 points

Problem Statement

Given are strings S and T. Consider changing S to T by repeating the operation below. Find the minimum number of operations required to do so.

Operation: Choose one character of S and replace it with a different character.

Constraints

S and T have lengths between 1 and 2\times 10^5 (inclusive).

S and T consists of lowercase English letters.

S and T have equal lengths.

Input

Input is given from Standard Input in the following format:

S
T

Output

Print the answer.

Sample Input 1

cupofcoffee
cupofhottea

Sample Output 1

4

We can achieve the objective in four operations, such as the following:

First, replace the sixth character c with h.

Second, replace the eighth character f with t.

Third, replace the ninth character f with t.

Fourth, replace the eleventh character e with a.

Sample Input 2

abcde
bcdea

Sample Output 2

5

Sample Input 3

apple
apple

Sample Output 3

0

No operations may be needed to achieve the objective.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

///mAAria...
#include<bits/stdc++.h>
 
using namespace std;
 
#define int long long
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define all(v) v.begin(),v.end()
#define allr(v) v.begin(),v.end()
#define ld long double
#define ff first
#define ss second
#define pb push_back
#define pii pair<int,int>
#define pi 3.1415926535897932384626433832
#define mod 1000000007
//#define mod 998244353
#define inf 1000000000
#define pr1(a) cout << a << endl
#define pr2(a,b)cout << a <<  " " <<b<< endl
#define pr3(a,b,c) cout << a <<" " << b << " " << c << endl

const long double PI = acos(-1);

int powm(int a, int b) {
    int res=1;
    while(b) {
        if(b&1)
            res=(res*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return res;
}

int gcd(int a, int b){
    if(b==0)return a;
    return gcd(b,a%b);
}

bool p(int n){
    return ((n&(n-1))==0);

}
void solve(){
    
    string s, t;
    cin >> s >> t;

    int ans = 0;

    for(int i = 0; i<s.size(); i++){
        if(s[i]!=t[i])ans++;
    }
    cout << ans<< endl;
   
} 

signed main() 
{   
    IOS;
    int t = 1;
    // cin >> t;

    for(int i = 1;i<=t;i++){
        solve();
    }
    return 0;
}