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.InputStreamReader;
import java.lang.reflect.Array;
import java.util.*;

public class Main {public static void main(String[] args) throws Exception {new Solve();}}

class Solve { public Solve() throws Exception{solve();}
    void solve() throws Exception {
        int x = NI();
        int y = NI();
        int a = NI();
        int b = NI();
        int lcm = lcm(x, y);
        int m = (int)Math.ceil((double)a / lcm) * lcm;
        System.out.println(m > b ? 0 : ((b - m) / lcm + 1));
    }

    int lcm(int x, int y) {
        return x * y / gcd(min(x, y), max(x, y));
    }

    int gcd(int x, int y) {
        return x * y == 0 ? x + y : gcd(y % x, x);
    }

    void inInt(int[]a,int n)throws Exception{for(int i=0;i<n;i++)a[i]=NI();}
    void inLong(long[]a,int n)throws Exception{for(int i=0;i<n;i++)a[i]=NL();}
    BufferedReader stdin=new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st=new StringTokenizer("");
    int min(int i1,int i2){return i1<i2?i1:i2;}
    long min(long i1,long i2){return i1<i2?i1:i2;}
    int max(int i1,int i2){return i1>i2?i1:i2;}
    long max(long i1,long i2){return i1>i2?i1:i2;}
    String NS()throws Exception{while(!st.hasMoreTokens())st=new StringTokenizer(stdin.readLine());return st.nextToken();}
    String NLn()throws Exception{return stdin.readLine();}
    int NI()throws Exception{return Integer.parseInt(NS());}
    long NL()throws Exception{return Long.parseLong(NS());}
    double ND()throws Exception{return Double.parseDouble(NS());}
    int abs(int x){return x<0?-x:x;}
    long abs(long x){return x<0?-x:x;}
    int mod(int x, int mod){return(x+mod)%mod;}
}
class ArrayUtils{public static void mergeSort(long[]a){long[]src=a.clone();mergeSortInt(src, a, 0, a.length);}
    public static void mergeSortInt(long[]src,long[]dest,int low,int high) {
        int length = high - low;
        if(length == 1) return;
        if (length == 2) {
            if(dest[low] > dest[high-1]){long t = dest[low];dest[low] = dest[high-1];dest[high-1] = t;}
            return;}
        int mid=(low+high)>>>1;
        mergeSortInt(dest,src,low,mid);
        mergeSortInt(dest,src,mid,high);
        for(int i=low,p=low,q=mid;i<high;i++)if(q>=high||p<mid&&src[p]<=src[q])dest[i]=src[p++];else dest[i] = src[q++];}}