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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toStr(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
const double EPS = 1e-12;
const double PI = acos(-1.0);
const int INF = INT_MAX / 10;
struct unionfind {
  vector<int> data, rank;
  unionfind(int size) {
    for (int i = (0); i <= ((size)-1); ++i) {
      data.push_back(i);
      rank.push_back(0);
    }
  }
  int find(int x) { return (data[x] == x) ? x : data[x] = find(data[x]); }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (rank[x] < rank[y]) {
      data[x] = y;
    } else {
      data[y] = x;
      if (rank[x] == rank[y]) rank[x]++;
    }
  }
  bool same(int x, int y) { return find(x) == find(y); }
};
int main() {
  vector<int> f(1001);
  f[0] = 1;
  f[1] = 2;
  for (int i = (2); i <= (1000); ++i) {
    f[i] = (f[i - 1] + f[i - 2]) % 1001;
  }
  int N, d;
  while (cin >> N >> d) {
    unionfind uf(N + 1);
    for (int i = (1); i <= (N); ++i) {
      for (int j = (i + 1); j <= (N); ++j) {
        if (abs(f[i] - f[j]) < d) {
          uf.unite(i, j);
        }
      }
    }
    int ans = 0;
    vector<int> memo(N, 0);
    for (int i = (0); i <= ((N)-1); ++i) {
      int root = uf.find(i);
      if (memo[root] == 0) {
        memo[root] = 1;
        ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
