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 <cstdio>
#include <vector>
#include <cmath>
#include <queue>

#define N 1001

using namespace std;

int main() {
	int V, d;
	while ( scanf("%d%d", &V, &d) != EOF ) {
		vector<int> v(V);
		for (int i = 0; i < V; ++i) {
			if (i == 0) v[i] = 2;
			else if (i == 1) v[i] = 3;
			else v[i] = (v[i-1] + v[i-2]) % N;
		}

		vector< vector<int> > edge(V);
		for (int i = 0; i < V; ++i) for (int j = i+1; j < V; ++j) {
			if ( abs(v[i] - v[j]) < d ) {
				edge[i].push_back(j);
				edge[j].push_back(i);
			}
		}

		int ans = 0;
		vector<bool> f(V, false);
		for (int s = 0; s < V; ++s) {
			if (!f[s]) {
				++ans;

				queue<int> Q; Q.push(s);
				while ( !Q.empty() ) {
					int now = Q.front(); Q.pop();
					if (f[now]) continue;
					f[now] = true;

					for (int i = 0; i < edge[now].size(); ++i) {
						Q.push( edge[now][i] );
					}
				}
			}
		}
		cout << ans << endl;
	}
}