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: Time Limit Exceeded)


#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#include <prettyprint.hpp>
#define debug(...)  cerr << "[" << #__VA_ARGS__ << "]: ", d_err(__VA_ARGS__);
#else
#define debug(...) 83;
#endif

void d_err() {
    cerr << endl;
}

template <typename H, typename... T>
void d_err(H h, T... t) {
    cerr << h << " ";
    d_err(t...);
}

template <typename T>
void print(T x) {
    cout << x << endl;
}

#define ALL(x) (x).begin(), (x).end()
#define FOR(i, m, n) for (int i = (m); i < (n); ++i)
#define REVFOR(i, m, n) for (int i = (n - 1); i >= (m); --i)
#define REP(i, n) FOR(i, 0, n)
#define REVREP(i, n) REVFOR(i, 0, n)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define bcnt __builtin_popcountll

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<ll,ll> Pll;
typedef pair<int,int> Pin;

ll INF = 1e16;
int inf = 1e9;
ll MOD = 1e9+7;

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    
    ll N, M, K;
    cin >> N >> M >> K;

    queue<ll> A, B;
    REP(i, N) {
        ll a;
        cin >> a;
        A.push(a);
    }

    REP(i, M) {
        ll b; cin >> b;
        B.push(b);
    }

    ll ans = 0;
    ll k = 0;
    while(!A.empty() || !B.empty()) {
        ll t;
        if (A.empty()) {
            t = B.front(); B.pop();
        } else if (B.empty()) {
            t = A.front(); A.pop();
        } else {
            auto &q = A.front() < B.front() ? A : B;
            t = q.front(); q.pop();
        }

        if (k + t > K) break;

        k += t;
        ans++;
    }

    print(ans);

    
}