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
#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
#define loop(i,a,b) for(int i=a;i<b;i++)
#define rep(i,a) loop(i,0,a)
#define N 1001
using namespace std;

int main(){
	int V,d;
	while(cin>>V){
		cin>>d;
		int rank[V+1],parent[V+1],label[V+1];
		label[0]=1;
		label[1]=2;
		rep(i,V+1){
			rank[i]=1;
			parent[i]=i;
			if(i>=2)label[i]=(label[i-1]+label[i-2])%N;
		}
		loop(i,1,V+1){
			loop(j,1,V+1){
				if(abs(label[i]-label[j])<d){
					int now1=i,now2=j;
					while(parent[now1]!=now1){
						now1=parent[now1];
					}
					while(parent[now2]!=now2){
						now2=parent[now2];
					}
					if(rank[now1]>rank[now2]){
						parent[now2]=now1;
					}else if(rank[now1]<rank[now2]){
						parent[now1]=now2;
					}else{
						parent[now2]=now1;
						rank[now1]++;
					}
				}
			}
		}
		int cnt=0;
		loop(i,1,V+1){
			if(parent[i]==i)cnt++;
		}
		cout<<cnt<<endl;
	}
	return 0;
}