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 <bits/stdc++.h>

using namespace std;

struct UnionFind{
  vector<int> data;
  UnionFind(int n)
  {
    data.assign(n, -1);
  }
  int find(int k)
  {
    if(data[k] < 0) return k;
    return data[k] = find(data[k]);
  }
  void unite(int x, int y)
  {
    x = find(x), y = find(y);
    if(x == y) return;
    if(x > y) swap(x, y);
    data[x] += data[y];
    data[y] = x;
  }
  int size(int k)
  {
    return -data[find(k)];
  }
};

int memo[1010];

int fib(int v)
{
  if(v <= 0) return 1;
  if(~memo[v]) return memo[v];
  return memo[v] = (fib(v-1) + fib(v-2)) % 1001;
}

int main()
{
  int V, d;
  int F[1010];
  memset(memo, -1, sizeof(memo));
  for(int i = 0; i < 1000; i++) F[i] = fib(i+1);
  while(cin >> V >> d){
    UnionFind uf(V);
    for(int i = 0; i < V; i++){
      for(int j = i+1; j < V; j++){
	if(abs(F[i] - F[j]) < d) uf.unite(i, j);
      }
    }
    int cnt = 0;
    for(int i = 0; i < V; i++){
      if(uf.data[i] < 0) cnt++;
    }
    cout << cnt << endl;
  }
  return 0;
}