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.util.Scanner;

public class C {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		
		int a=in.nextInt();
		int b=in.nextInt();
		int l=in.nextInt();
		int r=in.nextInt();
		
		
		System.out.println(fun(a,b,l,r));
		
	}
	
	static int fun(int a,int b,int l,int r){
		if ((r-l+1)>2*(a+b))
			if (a<=b)
				return a+1;
			else
				return (2*a-b);
		
		int[] flag = new int[3*(a+b)+1];
		
		int temp=0;
		for (int i=1;i<=a;i++){
			flag[i]=i;
			temp=i;
		}
		
		for (int i=a+1;i<=a+b;i++){
			flag[i]=temp;
		}
		
		for (int i=a+b+1;i<=a+b+a;i++){
			int tc=1;
			while (in(flag,tc,a+b-a+1,i-1))
				tc++;
			flag[i]=tc;
			temp=flag[i];
		}
		for (int i=a+b+a+1;i<=a+b+a+b;i++){
			flag[i]=temp;
		}
		for (int i=2*(a+b)+1;i<=3*(a+b);i++){
			flag[i]=flag[i-(a+b)];
		}
		
		int[] flagc = new int[25];
		int templ = l % (a+b);
		if (templ==0) templ+=a+b;
		int tempr = r - l + templ;
		if (tempr<templ) tempr+=a+b;
		for (int i=templ;i<=tempr;i++){
			flagc[flag[i]]=1;
		}	
		
		int res=0;
		for (int i=1;i<=24;i++)
			res+=flagc[i];
		
		return res;
		
	}
	
	static boolean in(int[] a,int b,int l,int r){
		for (int i=l;i<=r;i++){
			if (a[i]==b)
				return true;
		}
		return false;
	}
	

}
