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.
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.StringTokenizer;

public class C
{
	static PrintWriter out = new PrintWriter(
			new BufferedOutputStream(System.out));
	static MyScanner sc = new MyScanner(System.in);
	static StringBuilder output = new StringBuilder();

	public static void main(String... args)
	{
		int a = sc.nextInt();
		int b = sc.nextInt();
		int l = sc.nextInt() - 1;
		int r = sc.nextInt();
		
		r -= l / (a + b) * (a + b);
		l -= l / (a + b) * (a + b);
		
		if (r > (a + b) * 3)
		{
			r -= r / (a + b) * (a + b);
			r += (a + b) * 3;
		}
		
		int[] seq = new int[r];
		
		for (int i = 0; i < r; i++)
		{
			seq[i] = i;
		}
		
		{
			int i = a;
			while (i < r)
			{
				if (i % (a + b) < a)
				{
					HashSet<Integer> used = new HashSet<Integer>();
					for (int j = i - 1; i - j <= a; j--)
					{
						used.add(seq[j]);
					}
					
					for (int j = 0; i % (a + b) != a && i < r; i++, j++)
					{
						while (used.contains(j))
						{
							j++;
						}
						used.add(j);
						
						seq[i] = j;
					}
				}
				else
				{
					seq[i] = seq[i - 1];
					i++;
				}
			}
		}
		
		int result;
		
		{
			HashSet<Integer> used = new HashSet<Integer>();
			
			for (int i = l; i < r; i++)
			{
				used.add(seq[i]);
			}
			
			result = used.size();
		}
		
		for(int i = a; i < a + b && i < r && b < r; i++)
		{
			seq[i] = seq[b];
		}
		
		{
			int i = a + b;
			while (i < r)
			{
				if (i % (a + b) < a)
				{
					HashSet<Integer> used = new HashSet<Integer>();
					for (int j = i - 1; i - j <= a; j--)
					{
						used.add(seq[j]);
					}
					
					for (int j = 0; i % (a + b) != a && i < r; i++, j++)
					{
						while (used.contains(j))
						{
							j++;
						}
						used.add(j);
						
						seq[i] = j;
					}
				}
				else
				{
					seq[i] = seq[i - 1];
					i++;
				}
			}
		}
		
		{
			HashSet<Integer> used = new HashSet<Integer>();
			
			for (int i = l; i < r; i++)
			{
				used.add(seq[i]);
			}
			
			result = Math.min(result, used.size());
		}
		

		
		output.append(result);
		

		eOP();
	}
	
	private static void eOP()
	{
		System.out.println(output);
		System.exit(0);
	}
	
	/**
	 * Flatfoot's Scanner with slight modifications.
	 * 
	 * @author <a href="http://codeforces.com/profile/Flatfoot">Flatfoot</a>
	 * @see <a href="http://codeforces.com/blog/entry/7018">Source</a>
	 */
	private static class MyScanner
	{
		BufferedReader br;
		StringTokenizer st;

		MyScanner(InputStream in)
		{
			this.br = new BufferedReader(new InputStreamReader(in));
		}

		String next()
		{
			while (this.st == null || !this.st.hasMoreElements())
			{
				try
				{
					this.st = new StringTokenizer(this.br.readLine());
				}
				catch (IOException e)
				{
					e.printStackTrace();
				}
			}
			return this.st.nextToken();
		}

		int nextInt()
		{
			return Integer.parseInt(this.next());
		}

		long nextLong()
		{
			return Long.parseLong(this.next());
		}

		double nextDouble()
		{
			return Double.parseDouble(this.next());
		}

		/**
		 * This method has been modified to be a bit more similar to Scanner's
		 * 
		 * @author <a href=
		 *         "http://codeforces.com/profile/Flatfoot">Flatfoot</a>,
		 *         <a href=
		 *         "http://codeforces.com/profile/JasonBaby">JasonBaby</a>
		 */
		String nextLine()
		{
			if (this.st == null || !this.st.hasMoreElements())
			{
				String str = "";
				try
				{
					str = this.br.readLine();
				}
				catch (IOException e)
				{
					e.printStackTrace();
				}
				return str;
			}
			else
			{
				StringBuilder str = new StringBuilder();
				while (this.st.hasMoreElements())
				{
					str.append(this.st.nextToken());
					str.append(" ");
				}
				str.deleteCharAt(str.length() - 1);
				return str.toString();
			}
		}

	}
}
