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)

/*Bismillahir Rahmanir Rahim*///{
#include <bits/stdc++.h>
#include <ext/numeric>
using namespace std; using __gnu_cxx::power;
using ll = long long; using ull = unsigned long long; using ld = long double;
const ll llinf = (1ll<<61)-1; const double eps = 1e-6, ldeps = 1e-9;
struct Dbg { static constexpr auto &os = cout;   // replace by cerr in interactive.
template<class C> static auto o(C *x)->decltype(os << *x, 0);
template<class C> static char o(...);
template<class C> typename enable_if<sizeof o<C>(0)!=1,Dbg&>::type operator<<(const C &x) {os<<x; return *this;}
template<class C> Dbg &dump(C b, C e) {
    for (struct{C i;int j;} i={(*this<<"\n[",b),0}; i.i!=e and i.j<26; i.i++, i.j++) *this<<", "+2*(i.i==b)<<*i.i;
    return *this << ']';
}
template<class C> typename enable_if<sizeof o<C>(0)==1, Dbg&>::type operator<<(const C &x) {
    return dump(begin(x), end(x));
}
template<class C, size_t X> typename enable_if<!is_same<C, const char>::value, Dbg&>::type operator<<(C (&x)[X]) {
    return dump(begin(x), end(x));
}
template<class B, class C> Dbg &operator<<(const pair<B, C> &x) {return *this<<'{'<<x.first<<", "<<x.second<<'}';}
Dbg &operator<<(ostream&(*x)(std::ostream&)) { os << x; return *this; }
} dbg;
void err(istringstream *iss) { delete iss; }
template<class C, class... Args> void err(istringstream *iss, const C &val, const Args &... args) {
    string name; *iss >> name; if (name.back()==',') name.pop_back();
    dbg << name << " = " << val << "; ", err(iss, args...);
}
#define bug(args...) dbg << __LINE__ << ": ", err(new istringstream(#args), args), dbg << endl
#define cbug(a, args...) if (a) bug(args);
#define ASSERT(a, o, b, args...) if (!((a)o(b))) bug(a, b, ##args), assert((a)o(b));
#define sl(a) int(strlen(a))
#define sz(a) int(a.size())
int T, Q, cn;//}
const int inf = 1000000007, mxn = 2000005;

char s[mxn], t[mxn];

signed main() { cin.tie(0)->sync_with_stdio(0); cin.exceptions(cin.failbit); cout.precision(11), cout.setf(ios::fixed);
    cin >> s >> t;
    int n = sl(s);
    int ans = 0;
    for (int i = 0; i < n; i++) {
        ans += s[i] != t[i];
    }
    cout << ans << '\n';
}