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 <cstdio>
#include <algorithm>
#include <utility>
#include <vector>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define fi first
#define se second
typedef long long ll;

int N,M,H,W,K,A,B;
string S;
string alpha = "abcdefghijklmnopqrstuvwxyz";
string ALPHA = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

int main() {
  class union_find{
        vector<long> parent, tree_size;
    public:
        union_find(unsigned long _n) : parent(_n), tree_size(_n, 1){
            for(unsigned long i = 0; i < _n; ++i)parent[i] = i;
        }
        ll find(ll x){
            while(parent[x] != x)x = parent[x] = parent[parent[x]];
            return x;
        }
        bool merge(ll a, ll b){
            a = find(a);
            b = find(b);
	    if(a==b) return false;
            if(tree_size[a] < tree_size[b])swap(a, b);
            tree_size[a] += tree_size[b];
            parent[b] = a;
            return true;
        }
	bool same(ll a, ll b){
	    a = find(a);
	    b = find(b);
	    return a == b;
	}
  };
  vector<ll> fib(1010);//1-indexed
  fib[0]=1;
  fib[1]=2;
  reps(i,2,1011) {
    fib[i] = fib[i-1]+fib[i-2];
    fib[i]%= 1001;
  }
  while(cin>>N>>M){
    ll ans = N;
    union_find uf(N+10);
    reps(i,1,N+1){
      reps(j,i+1,N+1){
	if(abs(fib[i]-fib[j])<M) ans -= uf.merge(i-1,j-1);
      }
    } cout<<ans<<endl;
  }
}

