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 <stdio.h>
#include <vector>
#include <algorithm>

#define N 1001

using namespace std;

vector <int> par;

void init(int);
int root(int);
void uni(int, int);
bool same(int, int);

int main(){
	int V, d;
	
	while(cin >> V >> d){
		
		init(V);
		vector <int> f(V);
		int temp1, temp2;
		for(int i = 0; i < V; i++){
			if(i == 0)
				f[i] = temp1 = 2;
			else if(i == 1)
				f[i] = temp2 = 3;
			else{
				f[i] = (temp1 + temp2) % N;
				temp1 = temp2;
				temp2 = f[i];
			}
		}
		
		sort(f.begin(), f.end());
		
		for(int i = 0; i < V - 1; i++)
			if(f[i + 1] - f[i] < d)
				uni(i, i + 1);
		
		int count = 1;
		for(int i = 0; i < V - 1; i++)
			if(root(i) != root(i + 1))
				count++;
		
		cout << count << endl;
	}
	
	return 0;
}

//初期化
void init(int n){
	par.resize(n);
	for(int i = 0; i < n; i++){
		par[i] = i;  //最初自分の集合の代表は自分
	}
}

//木の根を求める
int root(int x){
	if(par[x] == x)
		return x;
	else
		return par[x] = root(par[x]);
}

//ｘとｙの属する集合を併合
void uni(int x, int y){
	x = root(x);
	y = root(y);
	if(x == y)
		return;
	par[x] = y;
}

//ｘとｙが同じ集合に属するか否か
bool same(int x, int y){
	return root(x) == root(y);
}