Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.

All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.

Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").

The players take turns appending letters to string s. Mister B moves first.

Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.

Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.

Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.

Input

First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.

Output

Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.

Examples

Input

1 1 1 8


Output

2

Input

4 2 2 6


Output

3

Input

3 7 4 6


Output

1

Note

In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.

In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.

In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1.
/**
 * DA-IICT
 * Author : Savaliya Sagar
 */
import java.io.*;
import java.math.*;
import java.util.*;

import static java.util.Arrays.fill;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
import static java.util.Collections.sort;

public class C820 {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	void solve() {
		int a = ni();
		int b = ni();
		int l = ni();
		int r = ni();
		if(r-l==0){
			out.println(1);
			return;
		}
		String str ="";
		for(int i=0;i<a;i++){
			str+= (char)(i+'a');
		}
		for(int i=0;i<b;i++){
			str+=(char)(a-1+'a');
		}
		StringBuilder s = new StringBuilder();
		int x = l%(a+b);
		if(x==0)
			x = (a+b);
		s.append(str.substring(x-1));
		HashSet<Integer> h = new HashSet<>();
		for(int i=s.length()-1;i>=(s.length()-a) && i>=0;i--)
			h.add(s.charAt(i)-'a');
		int j = 0;
		for(int i=0;i<a;i++){
			while(h.contains(j)) j++;
			s.append((char)(j+'a'));
			j++;
		}
		for(int i=0;i<b;i++){
			s.append((char)(j-1+'a'));
		}
		
		h = new HashSet<>();
		for(int i=s.length()-1;i>=(s.length()-a) && i>=0;i--)
			h.add(s.charAt(i)-'a');
		j = 0;
		for(int i=0;i<a;i++){
			while(h.contains(j)) j++;
			s.append((char)(j+'a'));
			j++;
		}
		for(int i=0;i<b;i++){
			s.append((char)(j-1+'a'));
		}
		h.clear();
		for(int i=0;i<=r-l && i<s.length();i++){
			h.add(s.charAt(i)-'a');
		}
		out.println(h.size());
	}

	void run() throws Exception {
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if (!INPUT.isEmpty())
			tr(System.currentTimeMillis() - s + "ms");
	}

	public static void main(String[] args) throws Exception {
		new Thread(null, new Runnable() {
			public void run() {
				try {
					new C820().run();
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
		}, "1", 1 << 26).start();
	}

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}

	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b !=
										// ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n && !(isSpaceChar(b))) {
			buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private void tr(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}
}
