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 <stdio.h>
#include <stdlib.h>
#include <time.h>

#include <string.h>
#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
#include <bits/stdc++.h>
using namespace std;
 
const double TL = 1.5 * CLOCKS_PER_SEC;
mt19937 rng((unsigned int) chrono::steady_clock::now().time_since_epoch().count());
#define clr(x, y) memset(x, y, sizeof(x))
#define forn(i, n) for (int i = 0; i < n; i++)
#define LL long long
const double pi = acos(-1.0);

char a[200005], b[200005];

int main() {
#ifdef LOCAL
  auto start_clock = clock();
  freopen("in", "r", stdin);
#endif
  scanf("%s%s", a, b);
  int n = strlen(a);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      ans += 1;
    }
  }
  printf("%d\n", ans);


#ifdef LOCAL
 auto end_clock = clock();
 cout << "Time:" << (end_clock - start_clock) / CLOCKS_PER_SEC << endl;
#endif
  return 0;
}