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 <iostream>
#include <algorithm>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <iomanip>
#include <set>
#include <ctime>
#include <tuple>
#include <bitset>
#include <assert.h>
#include <deque>
#include <functional>
using namespace std;
typedef long long ll;

#define fi first
#define se second
#define rep(i, n) for (ll i = 0; i < n; i++)
#define debugA cerr << "AAAAA" << endl
#define debug_ cerr << "-------------" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define debug_vec(v)                   \
    cout << #v << endl;                \
    for (int i = 0; i < v.size(); i++) \
    {                                  \
        cout << v[i] << " ";           \
    }                                  \
    cout << endl;
#define debug_vec2(v)                         \
    cout << #v << endl;                       \
    for (int i = 0; i < v.size(); i++)        \
    {                                         \
        for (int j = 0; j < v[i].size(); j++) \
        {                                     \
            cout << v[i][j] << " ";           \
        }                                     \
        cout << endl;                         \
    }

template <typename T>
bool chmax(T &a, const T &b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}

template <typename T>
bool chmin(T &a, const T &b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}

using Graph = vector<vector<int>>;
using P = pair<int, int>;
using P1 = pair<int, pair<int, int>>; // クラスカル法とかで、{cost, {from, to}}的に使う。
const ll LINF = 1001002003004005006ll;
const ll INF = 1001001001;
const double pi = acos(-1);

int main()
{
    string s, t;
    cin >> s >> t;
    int ans = 0;
    for (int i = 0; i < s.size(); i++)
    {
        if (s[i] != t[i])
        {
            ans++;
        }
    }
    cout << ans << endl;

    return 0;
}