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 <bits/stdc++.h>
using namespace std;

#define for_(i,a,b) for(int i=(a);i<(b);++i)

int V, d;
vector< vector< int > > adj;
vector< bool > vis;
vector< int > F;

void dfs(int v) {
	vis[v] = true;
	for (int u : adj[v]) if (!vis[u]) dfs(u);
}

int main() {
	vector< int > F(1001);
	F[0] = 1; F[1] = 2;
	
	for_(i,2,1001) F[i] = (F[i - 1] + F[i - 2]) % 1001;
	
	while (cin >> V >> d) {
		adj.assign(V + 1, vector< int >());
		
		for_(u,1,V+1) for_(v,u+1,V+1) {
			if (abs(F[u] - F[v]) < d) {
				adj[u].push_back(v);
				adj[v].push_back(u);
			}
		}
		
		vis.assign(V + 1, false);
		
		int ans = 0;
		
		for_(v,1,V+1) {
			if (vis[v]) continue;
			dfs(v);
			++ans;
		}
		
		cout << ans << endl;
	}
}