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>

//{ START
using namespace std;
#define int int64_t
#define rep(i, a, n) for (int i = (a); i < (n); ++i)
#define reps(i, a, n) for (int i = (a); i > (n); --i)
#define arep(i, x) for (auto &&i : (x))
#define irep(i, x) for (auto i = (x).begin(); i != (x).end(); ++i)
#define rirep(i, x) for (auto i = (x).rbegin(); i != (x).rend(); ++i)
//降順はgreater<T>()
#define all(x) (x).begin(), (x).end()
#define rv(s) reverse((s).begin(), (s).end())
// gcd lcmはそのままok
#define gcd(a, b) __gcd(a, b)
#define bits(n) (1LL << (n))
#define pcnt(x) __builtin_popcountll(x)
//配列内等要素削除
#define unique(x) (x).erase(unique((x).begin(), (x).end()), (x).end())
#define Fixed(n) fixed << setprecision(n)
//総和
#define sowa(n) ((n) * ((n) + 1)) / 2
#define pb(x, a) (x).push_back(a)
#define fb(x, a) (x).push_front(a)
#define eb(x, a) (x).emplace_back(a)
#define cauto const auto &
using P = pair<int, int>;
using Graph = vector<vector<P>>;
template <class T>  //昇順
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <class T>  //降順
using max_heap = priority_queue<T>;
template <class A, class B>
using umap = unordered_map<A, B>;
template <class A>
using uset = unordered_set<A>;
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {  //多次元初期化
  std::fill((T *)array, (T *)(array + N), val);
}
template <class A, class B>
bool chmax(A &a, const B &b) {  //最大値更新 返り値はbool
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class A, class B>
bool chmin(A &a, const B &b) {  //最小値更新 返り値はbool
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
constexpr int INF = 0x3f3f3f3f;
constexpr int LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr int mod1 = 1e9 + 7;
constexpr int mod2 = 998244353;
//} END

signed main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);

  string s, t;
  cin >> s >> t;

  int ans = 0;
  rep(i, 0, s.size()) ans += s[i] != t[i];
  cout << ans << '\n';
  return 0;
}