Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on. 

Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.

After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question. 

Input

The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).

Output

Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.

Examples

Input

2 3 6 18


Output

3

Note

Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. 
import java.io.*;
import java.math.*;
import java.util.*;

//Codeforces
public class MainCodeforces1 {
	private static MyScanner in;
	private static PrintStream out;
	private static boolean LOCAL_TEST = false;

	private static void solve() throws IOException
	{
		long x = in.nextLong();
		long y = in.nextLong();
		long a = in.nextLong();
		long b = in.nextLong();
		long lcm = LCM(x, y);
		long first = -1;
		for (long i = a; i <= b; i++) {
			if (i % lcm == 0) {
				first = i;
				break;
			}
		}
		if (first == -1) {
			out.println(0);
			return;
		}
		out.println(((b - first) / lcm) + 1);
	}

	public static long LCM(long a, long b) {
		return a / GCD(a, b) * b;
	}

	public static long GCD(long a, long b) {
		long rem = a % b;
		while (rem != 0) {
			a = b;
			b = rem;
			rem = a % b;
		}
		return b;
	}

	public static void main(String[] args) throws IOException {
		// helpers for input/output
		out = System.out;
		try {
			String cname = System.getenv("COMPUTERNAME");
			if (!cname.equals(""))
				LOCAL_TEST = true;
		} catch (Exception e) {
		}
		if (LOCAL_TEST) {
			in = new MyScanner("E:\\zin.txt");
		}
		else {
			boolean usingFileForIO = false;
			if (usingFileForIO) {
				// using input.txt and output.txt as I/O
				in = new MyScanner("input.txt");
				out = new PrintStream("output.txt");
			}
			else {
				in = new MyScanner();
				out = System.out;
			}
		}

		solve();
	}

	// =====================================
	static class MyScanner {
		BufferedReader bufReader;
		StringTokenizer strTok;

		public MyScanner() throws IOException
		{
			bufReader = new BufferedReader(new InputStreamReader(System.in));
			strTok = new StringTokenizer("");
		}

		public MyScanner(String inputFile) throws IOException {
			bufReader = new BufferedReader(new InputStreamReader(new FileInputStream(
					inputFile)));
			strTok = new StringTokenizer("");
		}

		String GetNextToken() throws IOException {
			if (!strTok.hasMoreTokens())
				strTok = new StringTokenizer(bufReader.readLine());
			return strTok.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.valueOf(GetNextToken());
		}

		public long nextLong() throws IOException {
			return Long.valueOf(GetNextToken());
		}

		public double nextDouble() throws IOException {
			return Double.valueOf(GetNextToken());
		}

		public String nextString() throws IOException {
			return GetNextToken();
		}

		public String nextLine() throws IOException {
			return bufReader.readLine();
		}

		public int countTokens() {
			return strTok.countTokens();
		}

		public boolean hasMoreTokens() {
			return strTok.hasMoreTokens();
		}
	}

}