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.*;
import java.util.*;

import static java.lang.Math.*;

public class Main extends PrintWriter {
    BufferedReader in;
    StringTokenizer stok;
    final Random rand = new Random(31);
    final int inf = (int) 1e9;
    final long linf = (long) 1e18;

    int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    public void solve() throws IOException {
        int x = nextInt();
        int y = nextInt();
        int a = nextInt();
        int b = nextInt();
        int lcm = x * y / gcd(x, y);
        println(b / lcm - (a - 1) / lcm);
    }

    public void run() {
        try {
            solve();
            close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(abs(-1));
        }
    }

    Main() throws IOException {
        super(System.out);
        in = new BufferedReader(new InputStreamReader(System.in));
    }

    Main(String s) throws IOException {
        super("".equals(s) ? "output.txt" : (s + ".out"));
        in = new BufferedReader(new FileReader("".equals(s) ? "input.txt" : (s + ".in")));
    }

    public static void main(String[] args) throws IOException {
        try {
            Locale.setDefault(Locale.US);
        } catch (Exception ignored) {
        }
        new Main().run();
    }

    String next() throws IOException {
        while (stok == null || !stok.hasMoreTokens()) {
            String s = in.readLine();
            if (s == null) {
                return null;
            }
            stok = new StringTokenizer(s);
        }
        return stok.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    int[] nextIntArray(int len) throws IOException {
        int[] a = new int[len];
        for (int i = 0; i < len; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    void shuffle(int[] a) {
        for (int i = 1; i < a.length; i++) {
            int x = rand.nextInt(i + 1);
            int t = a[i];
            a[i] = a[x];
            a[x] = t;
        }
    }

    boolean nextPerm(int[] p) {
        for (int a = p.length - 2; a >= 0; --a)
            if (p[a] < p[a + 1])
                for (int b = p.length - 1; ; --b)
                    if (p[b] > p[a]) {
                        int t = p[a];
                        p[a] = p[b];
                        p[b] = t;
                        for (++a, b = p.length - 1; a < b; ++a, --b) {
                            t = p[a];
                            p[a] = p[b];
                            p[b] = t;
                        }
                        return true;
                    }
        return false;
    }

    <T> List<T>[] createAdjacencyList(int countVertex) {
        List<T>[] res = new List[countVertex];
        for (int i = 0; i < countVertex; i++) {
            res[i] = new ArrayList<T>();
        }
        return res;
    }
}