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<stdio.h>
#include<algorithm>
using namespace std;
int UF[1001];
int FIND(int a){
	if(UF[a]<0)return a;
	return UF[a]=FIND(UF[a]);
}
void UNION(int a,int b){
	a=FIND(a);
	b=FIND(b);
	if(a==b)return;
	UF[a]+=UF[b];
	UF[b]=a;
}
int fib[1000];
int ABS(int a){
	return max(a,-a);
}
int main(){
	int a,b;
	fib[0]=2;
	fib[1]=3;
	for(int i=2;i<1000;i++)fib[i]=(fib[i-1]+fib[i-2])%1001;
	while(~scanf("%d%d",&a,&b)){
		for(int i=0;i<1001;i++)UF[i]=-1;
		for(int i=0;i<a;i++){
			for(int j=i+1;j<a;j++)if(ABS(fib[i]-fib[j])<b)UNION(i,j);
		}
		int ret=0;
		for(int i=0;i<a;i++)if(UF[i]<0)ret++;
		printf("%d\n",ret);
	}
}