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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class A819 {
	static int count(int[] cc, int l, int r) {
		boolean[] used = new boolean[24];
		int cnt = 0;
		for (int i = l; i <= r; i++) {
			int c = cc[i];
			if (!used[c]) {
				used[c] = true;
				cnt++;
			}
		}
		return cnt;
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int a = Integer.parseInt(st.nextToken());
		int b = Integer.parseInt(st.nextToken());
		int l = Integer.parseInt(st.nextToken()) - 1;
		int r = Integer.parseInt(st.nextToken()) - 1;
		int ab = a + b;
		int m = 4;	//	ab ab ab ab
		int[] cc = new int[ab * m];
		int h = l / ab;
		l -= h * ab;
		r -= h * ab;
		h = (r - l + 1) / ab;
		if (h > 2)
			r -= (h - 2) * ab;
		int d = a > b ? a - b : 1;
		for (h = 0; h < m; h++) {
			for (int k = 0; k < a - d; k++)
				cc[(a + b) * h + k] = k;
			for (int k = 0; k < d; k++)
				cc[(a + b) * h + a - d + k] = h % 2 == 0 ? a - d + k : a + k;
		}
		int min = a + a;
		for (int k0 = 1; k0 <= d; k0++) {
			for (int k = 0; k < b; k++)
				cc[(a + b) * 0 + a + k] = cc[(a + b) * 0 + a - k0];
			for (int k1 = 1; k1 <= d; k1++) {
				for (int k = 0; k < b; k++)
					cc[(a + b) * 1 + a + k] = cc[(a + b) * 1 + a - k1];
				for (int k2 = 1; k2 <= d; k2++) {
					for (int k = 0; k < b; k++)
						cc[(a + b) * 2 + a + k] = cc[(a + b) * 2 + a - k2];
					for (int k3 = 1; k3 <= d; k3++) {
						for (int k = 0; k < b; k++)
							cc[(a + b) * 3 + a + k] = cc[(a + b) * 3 + a - k3];
						min = Math.min(min, count(cc, l, r));
					}
				}
			}
		}
		System.out.println(min);
	}
}
