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 java.lang.Math;
import java.math.BigInteger;

public class Problem {

	public static void main(String[] args) throws IOException {

		new Problem().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer tok;
	Random rnd = new Random();
	static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;

	Problem() throws FileNotFoundException {

		if (ONLINE_JUDGE) {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
		} else {
			in = new BufferedReader(new FileReader("input.txt"));
			out = new PrintWriter("output.txt");
		}
		tok = new StringTokenizer("");
	}

	List<Integer>[] graph;
	boolean[] used1;
	boolean[] used2;
	int[] ans;
	int[] a;
	int[] countDiv;
	ArrayList<Integer> divs = new ArrayList<>();;

	void solve() throws IOException {
		int x = rI();
		int y = rI();
		int a = rI();
		int b = rI();
		int lcm = (x*y)/gcd(x,y);
		int new_b = b-b%lcm;
		int new_a = a+lcm-((a%lcm)==0?lcm:(a%lcm));
		int ans = (new_b-new_a)/lcm+1;
		out.println(ans);
	}

	// ======================================================
	class SegmentTree {

		int[] tree;

		SegmentTree(int n) {
			tree = new int[n];
		}

		void build(int[] a, int v, int curL, int curR) {
			if (curL == curR) {
				this.tree[v] = a[curL];
			} else {
				int mid = (curL + curR) / 2;
				build(a, v * 2, curL, mid);
				build(a, v * 2 + 1, mid + 1, curR);
				this.tree[v] = this.tree[v * 2] + this.tree[v * 2 + 1];
			}
		}

	}

	class Dsu {
		int[] parents;
		int countUnions;

		Dsu(int size) {
			this.parents = new int[size];
			for (int i = 0; i < size; ++i) {
				parents[i] = i;
			}
			this.countUnions = size;
		}

		int get(int v) {
			if (v == parents[v])
				return v;
			return parents[v] = get(parents[v]);
		}

		boolean union(int a, int b) {
			a = get(a);
			b = get(b);
			if (a == b) {
				return false;
			}
			if (rnd.nextBoolean()) {
				parents[a] = b;
			} else {
				parents[b] = a;
			}
			this.countUnions--;
			return true;
		}
	}

	// ======================================================
	void run() throws IOException {
		solve();
		out.close();
	}

	char[] reverseCharArray(char[] arr) {
		char[] ans = new char[arr.length];
		for (int i = 0; i < arr.length; ++i) {
			ans[i] = arr[arr.length - i - 1];
		}
		return ans;
	}

	int sqrt(double m) {
		int l = 0;
		int r = 1000000000 + 9;
		int i = 1000;
		while (r - l > 1) {
			int mid = (r + l) / 2;
			if (mid * mid > m) {
				r = mid;
			} else {
				l = mid;
			}
		}
		return l;
	}

	int countPow(int m, int n) {
		int ans = 0;
		while (m % n == 0 && m > 0) {
			ans++;
			m /= n;
		}
		return ans;
	}

	long binPow(long a, long b) {
		if (b == 0) {
			return 1;
		}
		if (b % 2 == 1) {
			return a * binPow(a, b - 1);
		} else {
			long c = binPow(a, b / 2);
			return c * c;
		}

	}

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

	long pow(long x, long k) {
		long ans = 1;
		for (int i = 0; i < k; ++i) {
			ans *= x;
		}
		return ans;
	}

	// ////////////////////////////////////////////////////////////////////

	String delimiter = " ";

	String readLine() throws IOException {
		return in.readLine();
	}

	String rS() throws IOException {
		while (!tok.hasMoreTokens()) {
			String nextLine = readLine();
			if (null == nextLine)
				return null;

			tok = new StringTokenizer(nextLine);
		}

		return tok.nextToken(delimiter);
	}

	int rI() throws IOException {
		return Integer.parseInt(rS());
	}

	long rL() throws IOException {
		return Long.parseLong(rS());
	}

	int[] rA(int b) {
		int a[] = new int[b];
		for (int i = 0; i < b; i++) {
			try {
				a[i] = rI();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return a;
	}

	int[] readSortedIntArray(int size) throws IOException {
		Integer[] array = new Integer[size];

		for (int index = 0; index < size; ++index) {
			array[index] = rI();
		}
		Arrays.sort(array);

		int[] sortedArray = new int[size];
		for (int index = 0; index < size; ++index) {
			sortedArray[index] = array[index];
		}

		return sortedArray;
	}

	int[] sortedIntArray(int size, int[] array) throws IOException {
		for (int index = 0; index < size; ++index) {
			array[index] = rI();
		}
		Arrays.sort(array);

		int[] sortedArray = new int[size];
		for (int index = 0; index < size; ++index) {
			sortedArray[index] = array[index];
		}

		return sortedArray;
	}
}