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 <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;

struct UnionFind {
    vector<int> par;

    UnionFind(int v) : par(v, -1) { }

    int root(int x) {
        return par[x] < 0 ? x : par[x] = root(par[x]);
    }

    bool same(int x, int y) {
        return root(x) == root(y);
    }

    void unite(int x, int y) {
        x = root(x), y = root(y);
        if(x != y) {
            par[x] += par[y]; par[y] = x;
        }
    }

    int size(int x) {
        return -par[root(x)];
    }
};


int main(void){
  int f[1001],m=1001,V,d;
  f[0]=1;f[1]=2;
  for(int i=2;i<=1000;++i){f[i]=(f[i-1]+f[i-2])%m;}
  while(cin>>V>>d){
    UnionFind uf(V);
    for(int i=1;i<=V;++i){
      for(int j=1;j<=V;++j){
        if(abs(f[i]-f[j])<d)uf.unite(i-1,j-1);
      }
    }
    int cnt=0;
    for(int i=0;i<V;++i)cnt+=(uf.par[i] < 0)?1:0;
    // for(int i = 0; i < V; i++) cerr << f[i+1] << ' ' << uf.root(i) << endl;
    cout<<cnt<<endl;
  }
  return 0;
}