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 n;
  vector<int> r,p;
  UnionFind(){}
  UnionFind(int sz):n(sz),r(sz,1),p(sz,0){iota(p.begin(),p.end(),0);}
  int find(int x){
    return (x==p[x]?x:p[x]=find(p[x]));
  }
  bool same(int x,int y){
    return find(x)==find(y);
  }
  void unit(int x,int y){
    x=find(x);y=find(y);
    if(x==y) return;
    if(r[x]<r[y]) swap(x,y);
    r[x]+=r[y];
    p[y]=x;
  }
};
int F[1009];
void fib(){
  F[0]=2;
  F[1]=3;
  for(int i=2;i<1009;i++){
    F[i]=F[i-1]+F[i-2];
    F[i]%=1001;
  }
}
int main(){
  int n,d;
  fib();
  while(cin>>n>>d){
    UnionFind U(n);
    for(int i=0;i<n;i++){
      for(int j=i+1;j<n;j++){
        if(abs(F[i]-F[j])<d){
          U.unit(i,j);
        }
      }
    }
    set<int>s;
    for(int i=0;i<n;i++)s.insert(U.find(i));
    cout<<s.size()<<endl;
  }
}
