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 <cstdlib>
#include <set>
using namespace std;

const int N=1001;
const int SIZE = 1000 + 100;
int parent[SIZE];
int f[SIZE];

int cal(int n){
  if( f[n] > 0 ) return f[n];
  return f[n] = (cal(n-1) + cal(n-2)) % N;
}

int find(int n){
  if( parent[n]==n ) return n;
  return parent[n] = find(parent[n]);
}

int main(){
  f[0] = 1;
  f[1] = 2;
  for( int i=2;i<SIZE;i++ ) f[i] = -1;

  int v,d;
  while( cin>>v>>d ){
    for( int i=1;i<=v;i++ )
      parent[i] = i;

    for( int i=1;i<=v;i++ ){
      for( int j=i+1;j<=v;j++ )
	if( abs( cal(i) - cal(j) ) < d )
	  parent[find(j)] = find(i);
    }
    set<int> s;
    for( int i=1;i<=v;i++ )
      s.insert(find(i));

    cout << s.size() << endl;
  }

  return 0;
}