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 <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
const int INF=1<<29,MOD=1e9+7;

//0~10000の頂点のroot
int v[10010];
//初期化(n頂点バラバラ)
void init(int n){
	rep(i,n) v[i]=i;
}

//頂点aが属するrootを求める
int root(int a){
	if(v[a]==a) return a; //aはroot
	return (v[a]=root(v[a])); //aの親のroot
}

//頂点aとbが同一グループか?
bool is_same_set(int a,int b){
	return root(a)==root(b);
}

//頂点aとbを同一グループにまとめる
void unite(int a,int b){
	v[root(a)]=root(b);
}

int main(){
	int fib[1010]; fib[0]=2;fib[1]=3;
	rep(i,1000) fib[i+2]=(fib[i+1]+fib[i])%1001;
	int V,d;
	while(cin>>V>>d){
		init(V);
		rep(i,V){
			for(int j=i+1;j<V;j++){
				if(abs(fib[i]-fib[j])<d) unite(i,j);
			}
		}
		//rep(i,V) cout<<root(i)<<endl;
		int cnt=0;
		rep(i,V){
			//cout<<root(i)<<endl;
			if(root(i)==i) cnt++;
		}
		cout<<cnt<<endl;
	}
	return 0;
}

