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<cmath>
using namespace std;
int fib[1010];
int d,V;
int parent[1010];
void makefib(){
  fib[0]=2;fib[1]=3;
  for(int i=2;i<1010;i++)fib[i]=(fib[i-1]+fib[i-2])%1001;
}
int root(int x){
  return (parent[x]==x)?x:root(parent[x]);
}
bool sameset(int x,int y){
  return root(x)==root(y);
}
void union_set(int x,int y){
  parent[root(y)]=root(x);
}
int main(){
  makefib();
  while(cin>>V>>d){
    for(int i=0;i<V;i++)parent[i]=i;
    for(int i=0;i<V;i++){
      for(int j=0;j<V;j++){
        if(abs(fib[i]-fib[j])<d&&!sameset(i,j))union_set(i,j);
      }
    }
    int cnt=0;
    for(int i=0;i<V;i++)if(root(i)==i)cnt++;
    cout<<cnt<<endl;
  }
}