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: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define faster ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
#define ll long long
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define vi vector<int>
#define pii pair<int,int>
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define REP(i,a,b) for(ll i = a; i < b; i++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) (a * b) / gcd(a, b)
#define bins(v, x) binary_search(v.begin(), v.end(), x)
#define pi 2*acos(0.0)
const int INF = 1e7;
const int mod = 1e9 + 7;

ll modExp(ll base, ll exp, ll mod) {
   ll res = 1;
   while (exp > 0) {
    	if (exp % 2 == 1) {
        	res= (res * base) % mod;
		}
      exp = exp >> 1;
      base = (base * base) % mod;
   }
   return res;
}

ll nCr(ll n,ll k){
   ll C[n + 1][k + 1];

   REP(i,0,n+1){
      REP(j,0,min(i,k)+1) {
         if(j==0 || j==i) {
            C[i][j] = 1;
         } else {
            C[i][j] = C[i-1][j-1] + C[i-1][j];
         }
      }
   }
   return C[n][k];
}

int main(){
   faster;
   //freopen("input.txt","r",stdin);
	 //freopen("output.txt","w",stdout);

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

   if(k <= a) p1(k);
   else if(k > a && k <= a + b) p1(a);
   else if(k > a + b && k <= a + b + c) p1(a + (k-a-b) * (-1));
}