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)

#include <bits/stdc++.h>
using namespace std;
using ll= long long;
using vi = vector<int>;
using vl= vector<ll>;
using vs=vector<string>;

ll gcd(ll a, ll b) { if (a == 0) return b; return gcd(b % a, a);}
ll max(ll a, ll b) {if (a > b) return a; else return b;}
ll min(ll a, ll b) {if (a < b) return a; else return b;}
ll power(ll a,ll b) {long long res = 1; while (b > 0) {if (b & 1)res = res * a;a = a * a;b >>= 1;} return res;}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int testcase=1;
    
    for(int p=0;p<testcase;p++){
        string s,t;
ll counter=0;
	cin>>s>>t;
	ll hashtable1[26]={0};
	ll hashtable2[26]={0};
	for(int i=0;i<s.size();i++){
	if(s[i]!=t[i])
	counter++;

}


cout<<counter;
        
        
        
        }

    return 0;
}