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
N = 1001

def init_fib
  fib = Array.new(1002)
  fib[0] = 1
  fib[1] = 2
  2.upto(N) do |i|
    fib[i] = (fib[i-1] + fib[i-2]) % N
  end
  fib
end

def root(tree, i)
  if tree[i] == i
    i
  else
    root(tree, tree[i])
  end
end

if __FILE__ == $0
  fib = init_fib
  STDIN.each_line do |input|
    count = 0
    v, d = input.split.map(&:to_i)
    tree = Array.new(v+1)

    # treeの初期化
    # tree[i] == i の場合、rootである
    0.upto(v) do |i|
      tree[i] = i
    end

    # 条件を満たすleafの統合
    # 常にrootに新しいleafをつける
    1.upto(v) do |i|
      (i+1).upto(v) do |j|
        tree[j] = root(tree, i) if (fib[j] - fib[i]).abs < d
      end
    end

    # rootの数を数える
    1.upto(v) do |i|
      count += 1 if root(tree, i) == i
    end

    # p tree
    puts count.to_s
  end
end