Problem:
Score : 200 points

Problem Statement

We have A cards, each of which has an integer 1 written on it. Similarly, we also have B cards with 0s and C cards with -1s.

We will pick up K among these cards. What is the maximum possible sum of the numbers written on the cards chosen?

Constraints

All values in input are integers.

0 \leq A, B, C

1 \leq K \leq A + B + C \leq 2 \times 10^9

Input

Input is given from Standard Input in the following format:

A B C K

Output

Print the maximum possible sum of the numbers written on the cards chosen.

Sample Input 1

2 1 1 3

Sample Output 1

2

Consider picking up two cards with 1s and one card with a 0.
In this case, the sum of the numbers written on the cards is 2, which is the maximum possible value.

Sample Input 2

1 2 3 4

Sample Output 2

0

Sample Input 3

2000000000 0 0 2000000000

Sample Output 3

2000000000

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <iomanip>
#include <cstdint>
#include <cmath>
#include <climits>
#include <cassert>
#include <string>
#include <string_view>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <numeric>
#include <functional>
#include <memory>

#define rep(i,n) for(int i = 0; i < (n); ++i)

using namespace std;

template<class T, class U>
istream& operator>>(istream& is, pair<T, U>& pair)
{
    is >> pair.first >> pair.second;
    return is;
}

template<class T>
istream& operator>>(istream& is, vector<T>& collection)
{
    for(auto& item : collection) is >> item;
    return is;
}

template <class T>
T peak_and_pop(std::queue<T>& q)
{
    assert(!q.empty());
    T t = move(q.front());
    q.pop();
    return t;
}

template <class T>
T peak_and_pop(std::priority_queue<T>& q)
{
    assert(!q.empty());
    T t = move(q.top());
    q.pop();
    return t;
}

template <class T>
T peak_and_pop(std::stack<T>& q)
{
    assert(!q.empty());
    T t = move(q.top());
    q.pop();
    return t;
}

void solution()
{
    int64_t a,b,c,k;
    cin >> a >> b >> c >> k;
    if(k <= a + b){
        cout << a << '\n';
    }else{
        cout << a - (k - a - b) << '\n';
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(10);
    solution();
    return 0;
}