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>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
using namespace std;
using ll = long long ;
using P = pair<int,int> ;
using pll = pair<long long,long long>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e17;
constexpr int MOD = 1001;
constexpr double PI = 3.14159265358979323846;


struct Unionfind{
    vector<int> parents;
    int n;
    Unionfind(int n):n(n),parents(n,-1){ }

    int find(int x){
        if(parents[x] < 0) return x;
        else return parents[x] = find(parents[x]);
    }

    void unite(int x,int y){
        x = find(x);
        y = find(y);

        if(x==y) return;
        if(parents[x] > parents[y]) swap(x,y);
        parents[x] += parents[y];
        parents[y] = x;
    }  

   int size(int x){
       return -parents[find(x)];
   }

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

   vector<int> members(int x){
       int root = find(x);
       vector<int> member;
       for(int i=0;i<n;i++){
           if(find(i)==root){
               member.push_back(i);
           }
       }
      return member;
  }

  int group_cnt(){
      int c = 0;
      rep(i,n){
          if(parents[i] < 0) ++c;
      }
      return c;
  }
};

int main(){
    int v,d;
    while(cin >> v >> d){
        vector<int> fib(v+2);
        fib[0] = fib[1] = 1;
        for(int i=2;i<v+2;i++){
            fib[i] = (fib[i-1] + fib[i-2])%MOD;
        }

        Unionfind uf(v);

        for(int i=2;i<v+2;i++){
            for(int j=i+1;j<v+2;j++){
                if(abs(fib[i]-fib[j])<d) uf.unite(i-2,j-2);
            }
        }

        cout << uf.group_cnt() << endl;
    }
    return 0;
}
