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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <cstring>
#include <climits>
using namespace std;
static const double EPS = 1e-5;
typedef long long ll;

set<int> fib;

int fi[1100];

int main(){
	fi[0]=1;
	fi[1]=2;
	for(int i=2;i<1100;i++){
		fi[i]=(fi[i-1]+fi[i-2])%1001;
	}
	
	int V,d;
	while(cin>>V>>d){
		fib.clear();
		for(int i=1;i<=V;i++){
			fib.insert(fi[i]);
		}
		
		int ans=1;
		int bef=*fib.begin();
		for(set<int>::iterator p=fib.begin();++p,p!=fib.end();){
			if(*p-bef>=d)++ans;
			bef=*p;
		}
		cout<<ans<<endl;
	}
}