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>
using namespace std;

#define int long long
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define all(v) (v).begin(),(v).end()
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define pb push_back
#define fi first
#define se second
template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

struct UF{
    vector<int>par,sz;
    void init(int n){
        par.resize(n);
        sz.resize(n);
        for(int i=0;i<n;i++){
            par[i]=i;
            sz[i]=1;
        }
    }
    int find(int x){
        return x==par[x]?x:par[x]=find(par[x]);
    }
    void unite(int x,int y){
        x=find(x);y=find(y);
        if(x==y)return;
        sz[x]+=sz[y];
        par[y]=x;
    }
    bool same(int x,int y){
        return find(x)==find(y);
    }
    int size(int x){
        return sz[find(x)];
    }
};

signed main(){
    int V,d;
    while(cin>>V>>d){
        int a=1,b=1;
        UF uf;uf.init(V);
        int F[1000];
        rep(i,V){
            F[i]=(a+b)%1001;

            b=a;a=F[i];
        }
        rep(i,V)rep(j,V)if(abs(F[i]-F[j])<d)uf.unite(i,j);
        int ans=0;
        rep(i,V)if(uf.find(i)==i)ans++;
        cout<<ans<<endl;
    }
    return 0;
}