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 <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <cstring>
#include <cassert>
#include <list>
#include <map>
#include <unordered_map>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <limits>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
#define left(x) (x << 1)
#define right(x) (x << 1) + 1
#define mid(l, r) ((l + r) >> 1)
#define mp make_pair
#define pb push_back
#define all(a) a.begin(), a.end()
#define debug(x)                          \
    {                                     \
        cerr << #x << " = " << x << "\n"; \
    }
#define debug2(x, y)                                                  \
    {                                                                 \
        cerr << #x << " = " << x << ", " << #y << " = " << y << "\n"; \
    }
#define debug3(x, y, z)                                                                           \
    {                                                                                             \
        cerr << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << "\n"; \
    }
#define debug4(x, y, z, w)                                                                                                    \
    {                                                                                                                         \
        cerr << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << ", " << #w << " = " << w << "\n"; \
    }
#define ss second
#define ff first
#define m0(x) memset(x, 0, sizeof(x))

template <typename T>
T min_(T a, T b)
{
    return a > b ? b : a;
}
template <typename T, typename... Ts>
T min_(T first, Ts... last) { return min_(first, min_(last...)); }
template <typename T>
T max_(T a, T b) { return a > b ? a : b; }
template <typename T, typename... Ts>
T max_(T first, Ts... last) { return max_(first, min_(last...)); }

template <class T, class S>
std::ostream &operator<<(std::ostream &os, const std::pair<T, S> &t)
{
    os << "(" << t.first << ", " << t.second << ")";
    return os;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v)
{
    out << "[";
    size_t last = v.size() - 1;
    for (size_t i = 0; i < v.size(); ++i)
    {
        out << v[i];
        if (i != last)
            out << ", ";
    }
    out << "]";
    return out;
}
const int mod = 1e9 + 7;

ll pwr(ll base, ll p)
{
    ll ans = 1;
    while (p)
    {
        if (p & 1)
            ans = (ans * base) % mod;
        base = (base * base) % mod;
        p /= 2;
    }
    return ans;
}
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }

const long double PI = (long double)(3.1415926535897932384626433832795);
const ll mx_ll = numeric_limits<ll>::max();
const int mx_int = numeric_limits<int>::max();

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    ll a, b, c, k;
    cin >> a >> b >> c >> k;

    ll fans = 0;
    if (k <= a)
    {
        fans = a;
    }
    else if (k > a && k <= (a + b))
    {
        fans = a;
    }
    else
    {
        fans = a - (k - (a + b));
    }
    cout << fans << endl;
    return 0;
}