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>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 

using namespace __gnu_pbds; 
using namespace std;
using   ll=long long;
using   ld=long double;

#define fast        	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ordered_set 	tree<pll, null_type,less<pll>, rb_tree_tag,tree_order_statistics_node_update>

#define endl        	"\n"
#define pb          	push_back
#define mp          	make_pair
#define F           	first
#define S           	second
#define int             long long int

#define ALL(v)      	v.begin(),v.end()
#define ALLR(v)     	v.rbegin(),v.rend()
#define pii         	3.14159265358979323
#define inf     		LLONG_MAX
#define ones(x)     	__builtin_popcount(x)
#define fill(a,b)   	memset(a,b,sizeof(a))
#define mod 			1000000007

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

ll mod_inverse(ll a) {
	return mod_pow(a , mod-2);
}

void solve() {
	
	string s , t;
	
	cin >> s >> t;
	
	int f[26] {};
	int ff[26] {};
	
	int sm = 0;
	
	for(int i = 0; i < s.size(); ++i) {
		if(s[i] != t[i]) sm++;
	}
	
	cout << sm;
}

signed main() {
    fast;
    
    int t = 1;
    
    //cin >> t;
    
    while(t--) {
        solve();
    }
 
    return 0;
}