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>
const int INF = 1001001001;
const int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1},
          dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
const int MOD = 1001;
int fib[1002];
class UnionFind {
 public:
  UnionFind() {}
  UnionFind(int n) { init(n); }
  void init(int n) {
    for (int i = (0); i < (n); i++) {
      parent[i] = i;
      rank[i] = 0;
    }
  }
  int find(int v) {
    if (v == parent[v]) {
      return v;
    }
    return parent[v] = find(parent[v]);
  }
  void unite(int x, int y) {
    if (rank[x] > rank[y]) {
      parent[y] = x;
    } else {
      parent[x] = y;
      if (rank[x] == rank[y]) {
        rank[y] += 1;
      }
    }
  }
  bool same(int u, int v) { return find(u) == find(v); }

 private:
  int parent[1002], rank[1002];
};
UnionFind uf;
int main() {
  fib[0] = 1;
  fib[1] = 2;
  for (int i = (2); i < (1002); i++) {
    fib[i] = (fib[i - 1] + fib[i - 2]) % MOD;
  }
  int V, d;
  while (std::cin >> V >> d, !std::cin.eof()) {
    uf.init(V + 1);
    for (int i = (1); i < (V + 1); i++) {
      for (int j = (i + 1); j < (V + 1); j++) {
        if (std::abs(fib[i] - fib[j]) < d) {
          uf.unite(i, j);
        }
      }
    }
    int res = 0;
    for (int i = (1); i < (V + 1); i++) {
      if (i == uf.find(i)) {
        res += 1;
      }
    }
    std::cout << res << std::endl;
  }
}
