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;

struct UnionFind{
    int num;
    vector<int> par,rank;
    UnionFind(int n):num(n),par(n),rank(n,1){
        iota(par.begin(),par.end(),0);
    }
    int root(int x){
        return (par[x]==x?x:par[x]=root(par[x]));
    }
    bool merge(int x, int y){
        x=root(x),y=root(y);
        if (x==y) return false;
        if (rank[x]<rank[y]) swap(x,y);
        par[y]=x;
        rank[x]+=rank[y];
        --num;
        return true;
    }
    bool same(int x, int y){return root(x)==root(y);}
    int size(int x){return rank[root(x)];}
    int count(){return num;}
};

const int N=1001;

int V,d;
int F[N];

void fibonacchi(){
    F[0]=2,F[1]=3;
    for (int i=2;i<N;++i) F[i]=(F[i-1]+F[i-2])%N;
}

void solve(){
    UnionFind uf(V);
    for (int i=0;i<V;++i){
        for (int j=i+1;j<V;++j){
            if (abs(F[i]-F[j])<d){
                uf.merge(i,j);
            }
        }
    }
    cout << uf.count() << '\n';
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    fibonacchi();
    while(cin >> V >> d){
        solve();
    }
}
