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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>

using namespace std;

#ifdef _MSC_VER
#define __typeof__ decltype

template <class T>
int __builtin_popcount(T n) { return n ? 1 + __builtin_popcount(n & (n - 1)) : 0; }
#endif

#define foreach(it, c) for (__typeof__((c).begin()) it=(c).begin(); it != (c).end(); ++it)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define popcount __builtin_popcount
	
#define rep(i, n) for (int i = 0; i < n; ++i)

template <class T> void max_swap(T& a, const T b) { a = max(a, b); }
template <class T> void min_swap(T& a, const T b) { a = min(a, b); }

const double EPS = 1e-10;
typedef long long ll;
typedef pair<int, int> pint;


int main()
{
	const int n = 1001;
	int f[n + 1];
	f[0] = 1, f[1] = 2;
	for (int i = 2; i <= n; ++i)
		f[i] = (f[i - 1] + f[i - 2]) % n;

	int v, d;
	while (~scanf("%d%d", &v, &d))
	{
		vector<int> s;
		for (int i = 1; i <= v; ++i)
			s.push_back(f[i]);
		sort(all(s));

		int res = 1;
		for (int i = 1; i < s.size(); ++i)
			if (s[i] - s[i - 1] >= d)
				++res;
		printf("%d\n", res);
	}
}