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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

//--- pB ---//
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
	public static void main (String[] args) throws java.lang.Exception {
		new Solution();
	}
}
class Solution {
	Reader reader;
	PrintWriter out;
	public Solution() {
		reader = new Reader();
		out = new PrintWriter(System.out);
		while (reader.hasNext()) {
			run_case();
		}
		out.flush();
	}

	private void run_case() {
		int a = reader.ni();
		int b = reader.ni();
		int c = reader.ni();
		int k = reader.ni();

		int res = 0;
		if (k <= a) {
			res = k;
		} else {
			res += a;
			k -= a;
			if (k <= b) {

			} else {
				k -= b;
				if (k <= c) {
					res -= k;
				}
			}
		}

		out.println(res);

		return;
	}
}


class Reader {
	BufferedReader br;
	StringTokenizer st;
	public Reader(){
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = new StringTokenizer(br.readLine());
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
	public String next() {
		if (st.hasMoreTokens()) return st.nextToken();
		try {
			st = new StringTokenizer(br.readLine());
		} catch (Exception e) {
			e.printStackTrace();
		}
		return st.nextToken();
	}
	public boolean hasNext() {
		try {
			if (!st.hasMoreTokens() && !br.ready()) return false;
			else return true;
		} catch (Exception e) {
			e.printStackTrace();
		}
		return false;
	}
	public int ni() {return Integer.parseInt(this.next());}
	public long nl() {return Long.parseLong(this.next());}
	public String ns() {return this.next();}
}