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 <set>
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class UnionFindTree{

private:
  // indexツづ個青板づ個親ツノツーツド
  vector<int> par;
  // indexツづーツ債ェツづつキツづゥツ姪伉づ個δ可δ督ク
  vector<int> rank;
  // ツ姪伉づ個催妥・ツ値
  int treeSize;
  
public:
  UnionFindTree(int initTreeSize = 1000){
     // ツ暗クツ青板づ与ツつヲツづァツづェツつスツ値ツづーツ催妥・ツ格ツ納ツ青板づつキツづゥUnionFindTreeツづ個催ャツ青ャ
    treeSize = initTreeSize;
    init();
  }
  
  
  void init(){
    
    for(int i = 0; i < treeSize; i++){
      par.push_back(i);
      rank.push_back(0);
    }
    
  }

  // ツ与ツつヲツづァツづェツつスツ青板つェツ格ツ納ツつウツづェツづつ「ツづゥツ姪伉づ個δ仰ーツトツづーツ探ツ催オ
  int find(int x){
    
    if(par[x] == x){
      return x;
    }
    else{
      return par[x] = find(par[x]);
    }
    
  }
  
  void unite(int x,int y){
    
    x = find(x);
    y = find(y);
    
    if(x == y)
      return;
    
    if(rank[x] < rank[y]){
      par[x] = y;
    }
    else{
      par[y] = x;
      if(rank[y] == rank[x]){
	rank[x]++;
      }
    }
    
  }
  
  bool same(int x,int y){
    return find(x) == find(y);
  }

};



int fib[1001];

int main(){

	int V,d;
	while(cin>>V>>d){
		int cnt=1;
		fib[0]=1;
		fib[1]=2;
		for(int i = 2; i < 1001; i++)
			fib[i]=(fib[i-1]+fib[i-2])%1001;
		sort(fib+1,fib+V+1);
		//UnionFindTree uft(1001);
		int prv=fib[1];
		for(int i = 2; i <= V; i++){
			if(abs(prv-fib[i])>=d){
				cnt++;
			}
			prv=fib[i];
		}
		cout<<cnt<<endl;
	}

	return 0;
}