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 <math.h>
using namespace std;

struct Matrix2x2{
	int a,b,c,d;
};
Matrix2x2 mult(Matrix2x2 A,Matrix2x2 B){
Matrix2x2 C={0};
	C.a=((A.a*B.a)%1001+(A.b*B.c)%1001)%1001;
	C.b=((A.a*B.b)%1001+(A.b*B.d)%1001)%1001;
	C.c=((A.c*B.a)%1001+(A.d*B.c)%1001)%1001;
	C.d=((A.c*B.b)%1001+(A.d*B.d)%1001)%1001;
	return C;
}

Matrix2x2 expt(Matrix2x2 A, int p){
	if(p==1){
		return A;
	}
	else if(p%2==1){
		Matrix2x2 T=expt(A,p-1);
		return mult(A,T);
	}
	else{
		Matrix2x2 T=expt(A,p/2);
		return mult(T,T);
	}
}
int p[1010];
void init(int nn){
	
	for (int ii=0;ii<nn+1;ii++){
		p[ii]=ii;
	}
}
int root(int a){
	if(p[a]==a) return a;
	else return (	p[a]=root(p[a]));
}
void unite (int a,int b){
	p[root(a)]=root(b);
}
int main() {
int f[1010];
for(int k=0;k<1010;k++){
if(k==0){
	f[k]=1;
}
else if(k==1){
	f[k]=2;
}
else{
Matrix2x2 A={1,1,1,0};
Matrix2x2 G=expt(A,k-1);
f[k-2]=G.a;
}
}
int v,d;
while(cin>>v>>d){
	int j=0;
	init (v);
	for(int i=1; i<=v;i++){
		for(int j=i+1;j<=v;j++){
			if(abs(f[i]-f[j])<d){
				unite(i,j);
			}
		}
	}
for (int r=1;r<=v;r++){
	if(p[r]==r){
		j=j+1;
	}
}
cout<<j<<endl;
}
	return 0;
}