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;
#define fastIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
typedef long long int lli;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef vector<pll> vop;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<int> vi;
typedef pair<int, int> pii;
#define REP(i,a,b) for( int i=a;i<b;i++)
#define REPR(i,a,b) for(int i=a;i>=b;i--)
#define adb(a) for(auto i:a) cout << i << " "; cout << endl
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define all(x) (x).begin(),(x).end()
#define endl "\n"
#define mem(a,b) memset(a,b,sizeof(a))
#define mod 1000000007
#define PI 3.14159265358979323846
const ll N = 100005;

struct LessSecond
{
    template<typename T, typename U>
    bool operator()(const std::pair<T, U> &x, const std::pair<T, U> &y) const
    {
        return x.second < y.second;
    }
};

ll power(ll a, ll b, ll m) { ll ans = 1; a = a % m; while (b) { if (b & 1) ans = (ans * a) % m; b /= 2; a = (a * a) % m; } return ans; }

ll lcm(ll a, ll b) { return (a * b) / __gcd(a, b); }

ll modInverse(ll a, ll m) { ll g = __gcd(a, m); return power(a, m - 2, m); }


int main()
{


    fastIO;
    int test = 1;
    //cin >> test;
    while (test--)
    {
        string s, t;
        cin >> s >> t;
        int ans = 0;
        REP(i, 0, s.size())
        {
            if (s[i] != t[i])
                ans++;
        }
        cout << ans << endl;
    }
    return 0;
}