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)


//NAME-ADITYYA OJHA
                                               /*
                                                      GURU(MENTOR)-AARADHYA
                                               */
/*
 *     Heart beats fast
 *         Colors and promises
 *             How to be brave
 *                 How can I love when I am afraid to fall...
                                             (courtesy-amnesiac_dusk)
 */

#include <bits/stdc++.h>
#define ll long long int
#define forin for(int i=0;i<n;i++)
#define pb push_back
#define vx vector<ll> x
#define vy  vector<ll> y
#define fast  ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define  f   first
#define  s   second
#define mod 1000000007
using namespace std;

/*
check for corner cases(for n==1)
 check for int overflow
 run on some sample cases before submitting
*/

/*FERMAT THEOREM

       pow(x,m-1)%m=1 (x and m are co-prime)
*/
/*MODULAR INVERSE

        pow(x,-1)=pow(x,m-2) (x and m are co-prime)
*/

/*
ll  mode(ll x,ll n,ll mod)
{
    if(n==0)return 1%mod;

    ll u=mode(x,n/2,mod);
    u=(u*u)%mod;
    if(n & 1)u=(u*x)%mod;
}
*/


int main()
{


   string s,t;
   cin>>s>>t;
   ll cnt=0;
   for(int i=0;i<int(s.size());i++)
   {
       if(s[i]!=t[i])cnt++;
   }
cout<<cnt<<endl;











}