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;
template <class T>
ostream &operator<<(ostream &o, const vector<T> &v) {
  o << "{";
  for (int i = 0; i < (int)v.size(); i++) o << (i > 0 ? ", " : "") << v[i];
  o << "}";
  return o;
}
const int MAX = 2000;
const int MOD = 1001;
struct UnionFind {
  vector<int> par;
  vector<int> rank;
  UnionFind(int n) : par(n), rank(n, 1) {
    for (int i = 0; i <= n - 1; i++) par[i] = i;
  }
  int root(int x) {
    if (par[x] == x)
      return x;
    else
      return par[x] = root(par[x]);
  }
  void unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return;
    if (rank[x] > rank[y]) swap(x, y);
    par[x] = y;
    if (rank[x] == rank[y]) rank[y]++;
  }
  bool same(int x, int y) { return root(x) == root(y); }
};
int f[MAX];
void solve(void) {
  int V, d;
  f[0] = 1, f[1] = 2;
  for (int i = 0; i <= MOD - 1; i++) f[i + 2] = (f[i] + f[i + 1]) % MOD;
  while (cin >> V >> d) {
    UnionFind uf(V);
    for (int i = 1; i <= V; i++) {
      int j = i;
      while (abs(f[j] - f[j + 1]) < d) {
        uf.unite(j - 1, j);
        j++;
      }
    }
    set<int> s;
    for (int i = 1; i <= V; i++) s.insert(uf.root(i - 1));
    cout << s.size() << endl;
  }
}
int main(void) {
  solve();
  return 0;
}
