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 <vector>
#include <queue>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <cstdlib>
using namespace std;

#define rep(i,n) for(int i=0; i<n; i++)
#define repa(i,s,e) for(int i=s; i<=e; i++)

typedef vector<int> vi;

int v, d;
const int N = 1001;
int p[N];
int r[N];
int f[N];

int find(int x) {
	if(p[x] == x) {
		return x;
	} else {
		return p[x] = find(p[x]);
	}
}

void unite(int x, int y) {
	x = find(x);
	y = find(y);
	if(x == y) return;

	if(r[x] < r[y]) {
		p[x] = y;
	} else {
		p[y] = x;
		if(r[x] == r[y]) r[x]++;
	}
}

void uft() {
	repa(i,1,v) {
		p[i] = i;
		r[i] = 0;
	}

	repa(i,1,v) {
		repa(j,i,v) {
			if(abs(f[i] - f[j]) < d) unite(i, j);
		}
	}

	int ans = 0;
	vector<int> memo(v+1, 0);
	repa(i,1,v) {
		int root = find(i);
		if(memo[root] == 0) {
			memo[root] = 1;
			ans++;
		}
	}
	cout << ans << endl;
}

void dfs(int n, int g, vi& groups) {
	queue<int> que;
	que.push(n);

	while(!que.empty()) {
		int p = que.front();
		que.pop();

		if(groups[p] != 0) continue;

		groups[p] = g;
		repa(i,n+1,v) {
			if(groups[i] == 0 && abs(f[p] - f[i]) < d) {
				que.push(i);
			}
		}
	}
}

void solve() {
	int g = 1;
	vi groups(v+1, 0);
	repa(i,1,v) {
		if(groups[i] == 0) {
			dfs(i, g, groups);
			g++;
		}
	}
	cout << (g-1) << endl;
}

int main() {
	fill(f, f+N, 0);
	f[0] = 1;
	f[1] = 2;
	repa(i,2,N-1) {
		f[i] = (f[i-1] + f[i-2]) % N;
	}

	long t, elapsed;
	while(cin>>v>>d) {
		//t = clock();
		uft();
		//elapsed = clock() - t;
		//cout << "Time = " <<  elapsed / 1000.0 << endl;

		//t = clock();
		//solve();
		//elapsed = clock() - t;
		//cout << "Time = " <<  elapsed / 1000.0 << endl;
	}
}