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
def root a
	return a if @p[a] == a
	return (@p[a] = root(@p[a]))
end

def isSameSet a, b
	return root(a) == root(b)
end

def unite a, b
	@p[root(a)] = root(b)
end

@fib=[1,1]
n=0
def f n
	return if n >= 1010
	@fib.push @fib[-1]+@fib[-2]
	f n+1
end

f 1

while true
	arg=gets
	break if arg == nil
	v,d=arg.split(" ").map{|e|e.to_i}
	
	@nowFib = @fib[(2..v+1)]
	@p=[]
	v.times{|t|@p.push t}
	(v-1).times{|t|
		unite(t+1, t) if (@nowFib[t+1]-@nowFib[t]) < d
	}
	p @p.uniq.count
end