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 <set>

using namespace std;

void solve()
{
	int V, d;
	while(cin >> V >> d)
	{
		unsigned long long int f0 = 1;
		unsigned long long int f1 = 1;
		set<int> numbers;
		vector<unsigned long long int> fib;
		fib.push_back(f0);
		fib.push_back(f1);
		int count = 2;
		while(numbers.size() < V)
		{
			f0 = fib[count - 2];
			f1 = fib[count - 1];
			unsigned long long int f2 = f0 + f1;
			fib.push_back(f2);
			f2 %= 1001;
			if(numbers.find(f2) == numbers.end())
			{
				numbers.insert(f2);
			}
			++count;
		}
		set<int>::iterator it = numbers.begin();
		int before_num;
		int ans = 0;
		before_num = *it;
		++it;
		while(it != numbers.end())
		{
			if(*it - before_num >= d)
			{
				++ans;
			}
			before_num = *it;
			++it;
		}
		cout << ans + 1 << endl;
	}
}

int main()
{
	solve();
	return(0);
}