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<set>
using namespace std;

int nlist[1010];

int abs(int a){
  return ((a > 0) ?(a) : (-(a)));
}

int parent(int i){
  if(nlist[i] == i) return i;
  return (nlist[i]=parent(nlist[i]));
}

int main(){
  int v,d;
  int fib[1010];
  fib[0]=fib[1]=1;
  for(int i=2;i<1010;i++){
    fib[i] = (fib[i-1]+fib[i-2])%1001;
  }
  while(cin>>v>>d){
    set<int> s;
    for(int i=0;i<v;i++) nlist[i]=i;
    for(int i=0;i<v;i++){
      for(int j=i+1;j<v;j++){
	if(abs(fib[i+2]-fib[j+2]) < d){
	  nlist[parent(j)] = parent(nlist[i]);
	}      
      }
    }
    for(int i=0;i<v;i++) s.insert(parent(i));
    cout<<s.size()<<endl;
  }
  return 0;
}