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-10;
const double PI = acos(-1.0);
const int INF = INT_MAX / 10;
const int MOD = 1000000007;
struct UnionFind {
  vector<int> data;
  UnionFind(int size) : data(size, -1) {}
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    return x != y;
  }
  bool find(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  int size(int x) { return -data[root(x)]; }
};
int main() {
  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 V, d;
  while (cin >> V >> d) {
    UnionFind uf(V + 1);
    for (int i = (1); i <= (V); ++i) {
      for (int j = (i + 1); j <= (V - 1); ++j) {
        if (abs(f[i] - f[j]) < d) {
          uf.unite(i, j);
          break;
        }
      }
    }
    set<int> ans;
    for (int i = (1); i <= (V); ++i) {
      ans.insert(uf.root(i));
    }
    cout << ans.size() << endl;
  }
  return 0;
}
