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;
const int MAX_N = 10000;
struct UnionFind {
  int par[MAX_N];
  int rank[MAX_N];
  void init(int n) {
    for (int i = 0; i < n; i++) {
      par[i] = i;
      rank[i] = 0;
    }
  }
  int find(int x) {
    if (par[x] == x)
      return x;
    else
      return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (rank[x] < rank[y])
      par[x] = y;
    else {
      par[y] = x;
      if (rank[x] == rank[y]) rank[x]++;
    }
  }
  bool same(int x, int y) { return find(x) == find(y); }
};
int fib(int x) {
  if (x == 0 || x == -1) return 1;
  return fib(x - 1) + fib(x - 2);
}
int main() {
  int V, d;
  const int N = 1001;
  while (1) {
    cin >> V >> d;
    UnionFind uf;
    uf.init(V);
    for (int i = 0; i < V; i++) {
      for (int j = 0; j < V; j++) {
        int x;
        x = ((fib(i) % N) - (fib(j) % N) > 0) ? (fib(i) % N) - (fib(j) % N)
                                              : (fib(j) % N) - (fib(i) % N);
        if (x < d) uf.unite(i, j);
      }
    }
    set<int> a;
    for (int i = 0; i < V; i++) a.insert(uf.find(i));
    cout << a.size() + 1 << endl;
  }
}
