Fibonacci number f(i) appear in a variety of puzzles in nature and math, including packing problems, family trees or Pythagorean triangles. They obey the rule f(i) = f(i - 1) + f(i - 2), where we set f(0) = 1 = f(-1).

Let V and d be two certain positive integers and be N ≡ 1001 a constant. Consider a set of V nodes, each node i having a Fibonacci label F[i] = (f(i) mod N) assigned for i = 1,..., V ≤ N. If |F(i) - F(j)| < d, then the nodes i and j are connected.

Given V and d, how many connected subsets of nodes will you obtain?

<image>

Figure 1: There are 4 connected subsets for V = 20 and d = 100.

Constraints

* 1 ≤ V ≤ 1000
* 1 ≤ d ≤ 150

Input

Each data set is defined as one line with two integers as follows:

Line 1: Number of nodes V and the distance d.

Input includes several data sets (i.e., several lines). The number of dat sets is less than or equal to 50.

Output

Output line contains one integer - the number of connected subsets - for each input line.

Example

Input

5 5
50 1
13 13


Output

2
50
8
import java.util.Scanner;

public class Main {
	
	static int[] par=new int[1001];
	static void init(int n) {
		for(int i=0; i<=n; i++) {
			par[i]=i;
		}
	}
	
	static int root(int x) {
		if(par[x]==x) return x;
		return par[x]=root(par[x]);
	}
	
	static void unite(int x, int y) {
		par[root(x)]=root(y);
	}
	
	public static void main(String[] args) {
		try(Scanner sc = new Scanner(System.in)){
			int[] F=new int[1001];
			for(int i=2; i<=1000; i++) {
				F[0]=1;
				F[1]=2;
				F[i]=F[i-2]+F[i-1];
				F[i]%=1001;
			}
			while(sc.hasNext()) {
				int V=sc.nextInt();
				int d=sc.nextInt();
				init(V);
				for(int i=1; i<=V; i++) {
					for(int j=i+1; j<=V; j++) {
						if(Math.abs(F[j]-F[i])<d) {
							unite(i, j);
						}
					}
				}
				int count=0;
				for(int i=1; i<=V; i++) {
					if(par[i]==i) {
						count++;
					}
				}
				System.out.println(count);
			}
			
			
			
		}
	}
}
