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)

//g++ -std=c++17 -Wshadow -Wall -fsanitize=address -fsanitize=undefined -D_GLIBCXX_DEBUG -g PROGRAM -o OUTPUT -DLOCAL -lasan -lubsan
#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>

#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define alli(a, n, k) (a + k), (a + n + k)
#define FP(i, a, b, k) for (__typeof(a) i = a; i < b; i += k)
#define FS(i, a, b, k) for (__typeof(a) i = a; i > b; i -= k)
#define IT(it, a) for (__typeof(a.begin()) it = a.begin(); it != a.end(); ++it)
#define eps 1e-6
#define pi 3.141592653589793
using namespace std;
using namespace __gnu_pbds;

template <class T>
inline T gcd(T x, T y)
{
    if (!y)
        return x;
    return gcd(y, x % y);
}
typedef vector<int> VII;
typedef vector<ll> VLL;
typedef pair<int, int> PII;
typedef vector<pair<int, int>> VPII;
typedef vector<pair<int, PII>> VPPI;
const int MOD = 1e9 + 7;
const int INF = 1e9;


inline ll modulo(ll a, ll m)
{
    return (a % m + m) % m;
}

inline ll modInverse(ll a, ll m)
{
    assert(__gcd(a, m) == 1);
    ll m0 = m;
    ll y = 0, x = 1;

    if (m == 1)
        return 0;

    while (a > 1)
    {
        ll q = a / m;
        ll t = m;

        m = a % m, a = t;
        t = y;

        y = x - q * y;
        x = t;
    }

    if (x < 0)
        x += m0;

    return x;
}

inline ll modPow(ll x, ll y, ll m)
{ //x^y % m
    if (y == 0)
        return 1LL;
    else if (y == 1)
        return x;
    else
    {
        ll ans = modPow(x, y / 2, m) % m;
        if (y & 1)
        {
            return (((ans * ans) % m) * x) % m;
        }
        else
        {
            return (ans * ans) % m;
        }
    }
}

vector<string> split(char d,string s)
{
    stringstream ss(s);
    string t;
    vector<string> words;
    while(getline(ss,t,d))
    {
        words.pb(t);
    }

    return words;
}

int main(int argc, char *argv[])
{
#ifdef LOCAL
    auto start = chrono::high_resolution_clock::now();
#endif
    
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int ans = 0;
    string s, t;
    cin >> s >> t;
    for (int i = 0; i < s.size();++i)
    if(s[i]!=t[i])
        ans++;

    cout << ans;

#ifdef LOCAL
    auto end = chrono::high_resolution_clock::now();
    double time_taken =
        chrono::duration_cast<chrono::nanoseconds>(end - start).count();

    time_taken *= 1e-9;

    cerr << "\nTime taken by program is : " << fixed
         << time_taken << setprecision(9);
    cerr << " sec" << endl;
#endif
    return 0;
}