Problem: 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. 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class P340A_TheWall {
	private static BufferedReader br;
	private static PrintWriter pw;
	
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(System.out);
		String textLine = br.readLine();
		String[] splitted = textLine.split("\\s+");
		long x = Long.parseLong(splitted[0]);
		long y = Long.parseLong(splitted[1]);
		long a = Long.parseLong(splitted[2]);
		long b = Long.parseLong(splitted[3]);
		long lcmXY = x * y / euclideanGcd(x, y);
		
		if (lcmXY < a) {
			a = lcmXY * (a / lcmXY + (a % lcmXY != 0 ? 1 : 0));
		} else {
			a = lcmXY;
		}
		long answer = (b - a < 0 ? 0 : 1 + (b - a) / lcmXY);
		pw.print(answer);
		pw.close();
	}
	
    /**
     * Calculate GCD of a and b interpreted as unsigned integers.
     */
    private static long binaryGcd(long x, long y) {
    	long a = x, b = y;
        if (b == 0)
            return a;
        if (a == 0)
            return b;

        // Right shift a & b till their last bits equal to 1.
        int aZeros = Long.numberOfTrailingZeros(a);
        int bZeros = Long.numberOfTrailingZeros(b);
        a >>>= aZeros;
        b >>>= bZeros;

        int t = (aZeros <= bZeros ? aZeros : bZeros);

        while (a != b) {
            if (a > b) {  // a > b as unsigned
                a -= b;
                a >>>= Long.numberOfTrailingZeros(a);
            } else {
                b -= a;
                b >>>= Long.numberOfTrailingZeros(b);
            }
        }
        return a<<t;
    }

    private static long euclideanGcd(long a, long b) {
		long p = a, q = b; // using local var will be looped faster 
		while (q != 0) {
			long mod = p % q;
			p = q;
			q = mod;
		}
		return p;
	}
    
}