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;
import java.util.TreeSet;

public class Div2_421C {

	public static void main(String[] args) throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer inputData = new StringTokenizer(reader.readLine());
		int a = Integer.parseInt(inputData.nextToken());
		int b = Integer.parseInt(inputData.nextToken());
		long l = Long.parseLong(inputData.nextToken());
		long r = Long.parseLong(inputData.nextToken());

		if (r - l + 1 >= 2 * (a + b)) {
			if (a <= b) {
				System.out.println(a + 1);
			} else {
				System.out.println(2 * a - b);
			}
			return;
		}

		StringBuilder cString = new StringBuilder();
		for (int i = 0; i < a; i++) {
			cString.append((char) ('a' + i));
		}

		l--;
		r--;

		TreeSet<Integer> used = new TreeSet<Integer>();

		
		int ans = 26;
		for (int first = 0; first < 26; first++) {
			for (int second = a; second < 26; second++) {
				for (int i = 0; i < b; i++) {
					cString.append((char)('a' + first));
				}
				used.clear();
				for (char c : cString.substring(cString.length() - a, cString.length()).toCharArray()) {
					used.add(c - 'a');
				}
				int nxt = 0;
				for (int i = 0; i < a; i++) {
					while (used.contains(nxt)) {
						nxt++;
					}
					cString.append((char) ('a' + nxt));
					nxt++;
				}
				for (int i = 0; i < b; i++) {
					cString.append((char)('a' + second));
				}

				int len = cString.length();

				used.clear();
				for (int i = (int) (l % len); i != (r + 1) % len; i = (i + 1) % len) {
					used.add(cString.charAt(i) - 'a');
				}
				if(used.size() < ans){
					ans = used.size();
				}
				cString.delete(a, cString.length());
			}
		}

		System.out.println(ans);
	}

}