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 <cstdio>
#include <cstdlib>
#include <utility>
#include <typeinfo>
#include <cmath>
#include <numeric>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <stack>
#include <map>
#include <string>
#define REP(i,n) for(int i=0;i<n;i++)
typedef long long ll;
using namespace std;
typedef vector<ll> vl;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
typedef vector<string> vs;

int P[10010];
void init(int N){ 
	for (int i=0; i<N; ++i) P[i] = i;
}

int root(int a){
	if(P[a] == a)return a;
	return (P[a]=root(P[a]));
}

void unite(int a, int b) {
	P[root(a)]=root(b);
}

int main() 
{
	//配列を求めておく
	int fib[1002]={0};
	fib[0]=1;
	fib[1]=1;
	for(int i=2;i<1000+2;i++){
		fib[i]=(fib[i-2]+fib[i-1])%1001;
	}
	
	int n,m;
	while(cin>>n>>m)
	{
		init(n);
		REP(i,n)REP(j,n)if(max(fib[i+2]-fib[j+2],-fib[i+2]+fib[j+2])<m)unite(i,j);
		
		int ans(0);
		REP(i,n)if(i==P[i])ans++;
		
		cout<<ans<<endl;
	}
	return 0;
}