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<cstring>
#include<cmath>
#include<cassert>
#include<iostream>
#include<sstream>
#include<string>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<utility>
#include<numeric>
#include<algorithm>
#include<bitset>
#include<complex>
#include<stack>

using namespace std;

typedef long long Int;
typedef vector<int> vint;
typedef pair<int,int> pint;
typedef vector<string> vstring;
typedef vector<pint> vpint;

struct Edge{int to,from,cost;};

#define debug cout

template<class T> void pv(T a, T b) { for (T i = a; i != b; ++i) debug << *i << " "; debug << endl; }
template<class T> void chmin(T &t, T f) { if (t > f) t = f; }
template<class T> void chmax(T &t, T f) { if (t < f) t = f; }
int in() { int x; scanf("%d", &x); return x; }

#define rep(i,n) for(int i=0;i<(n);++i)
#define repn(i,m,n) for(int i=(m);i<=(n);++i)
#define repd(i,n) for(int i=(n)-1;i>=0;--i)
#define repnd(i,m,n) for(int i=(n);i>=(m);--i)
#define rep0(i,n) for(i=0;i<(n);++i)
#define repn0(i,m,n) for(i=(m);i<=(n);++i)
#define repd0(i,n) for(i=(n)-1;i>=0;--i)
#define repnd0(i,m,n) for(i=(n);i>=(m);--i)
#define all(n) n.begin(),n.end()
#define sz(n) ((int)(n).size())
#define IL for(;;)
#define MP make_pair
#define PB push_back
#define SS stringstream
#define X second
#define Y first
#define PUTLINE debug<<"LINE:"<<__LINE__<<endl;

const int INF = 2147483647/3;
const double EPS = 1e-10;
const double PI = acos(-1.0);

const int dx[]={1,-1,0,0,1,-1,1,-1,0};
const int dy[]={0,0,1,-1,1,-1,-1,1,0};

vint v;

int main() {
	int a,b,c,d,e;
	while(cin>>a>>b){
		v.clear();
		if(a<=0){
			cout<<0<<endl;
		}else if(a==1){
			cout<<1<<endl;
		}else{
			v.PB(2);
			v.PB(3);
			c=2;
			d=3;
			repn(i,3,a){
				v.PB(e=(c+d)%1001);
				c=d;
				d=e;
			}
			sort(all(v));
			int res=1;
			rep(i,sz(v)-1){
//debug<<v[i]<<" ";
				if(v[i+1]-v[i]>=b)++res;
			}
//debug<<v[sz(v)-1]<<endl;
			cout<<res<<endl;
		}
	}
	return 0;
}