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 <algorithm>
using namespace std;

const int MAX_N = 1001;
int f[MAX_N] = {2,3};

int par[MAX_N];
int rank[MAX_N];

// フィボナッチ数列の計算
void fib_init(){
	for(int i = 2 ; i < MAX_N ; i++ ){
		f[i] = f[i-1] + f[i-2];
	}
	for(int i = 0 ; i < MAX_N ; i++ ){
		f[i] = f[i] % MAX_N;
	}
}

// 初期化
void init(int n){
	for(int i=0 ; i < n ; i++ ){
		par[i] = i;
		rank[i] = 0;
	}	
}

// x の root を求める
int find(int x){
	if( par[x] == x ){
		return x;
	}else{
		return par[x] = find( par[x] );
	}
}

// x と y の属する集合の併合
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[x] == rank[y] )
			rank[x]++;
	}
}

// x と y が同じ集合に属するかどうか 
bool same(int x, int y){
	return find(x) == find(y);
}

int main(){
	fib_init();
	
	int v, d;
	while( cin >> v >> d ){
		init( v );
		int memo[MAX_N] = {0};
		
		// |f(i) - f(j)| < d となるとき i と j は同じ集合に属する
		for(int i=0 ; i < v ; i++ ){
			for(int j=i+1 ; j < v ; j++ ){
				if( abs( f[i] - f[j] ) < d ){
					unite( i , j );
				}
			}
		}
		
		// 集合の個数を数える
		int ans=0;
		for(int i=0 ; i < v ; i++ ){
			int root = find(i);
			if( memo[root] == 0 ){
				memo[root] = 1;
				ans++;
			}
		}
		// 出力
		cout << ans << endl;
	}
}