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 <vector>
#include <iostream>
#include <cstdlib>
#include <map>
#include <stack>
using namespace std;
template<typename T, typename S> vector<T>& operator<<(vector<T>& a, S b) { a.push_back(b); return a; }
#define rep(i,n) for(int i=0;i<n;i++)


int v,d;
bool graph[1001][1001];


int main(){
	int fib[1001];
	fib[0] = 2 , fib[1] = 3;
	for(int i=2;i<1001;i++){
		fib[i] = (fib[i-1]+fib[i-2])%1001;
	}
	while(cin >> v >> d){
		bool done[1001] = {0};
		int ret = 0;
		rep(i,1001)rep(j,1001)graph[i][j] = false;
		rep(i,v)rep(j,v){
				if(abs(fib[j]-fib[i]) < d)graph[i][j] = true;
		}
		rep(i,v){
			if(done[i])continue;
			else ret++;
			stack<int> S;
			S.push(i);
			done[i] = true;
			while(S.size()){
				int pos = S.top(); S.pop();
				rep(j,v){
					if(graph[pos][j] && !done[j]){
						S.push(j);
						done[j] = true;
					}
				}
			}
		}
		cout << ret << endl;
	}
}