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;
struct UnionFind {
  int n;
  vector<int> r, p;
  UnionFind() {}
  UnionFind(int sz) : n(sz), r(sz, 1), p(sz, 0) { iota(p.begin(), p.end(), 0); }
  int find(int x) { return (x == p[x] ? x : p[x] = find(p[x])); }
  bool same(int x, int y) { return find(x) == find(y); }
  void unit(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (r[x] < r[y]) swap(x, y);
    r[x] += r[y];
    p[y] = x;
  }
};
int F[1009];
void fib() {
  F[0] = 2;
  F[1] = 3;
  for (int i = 2; i < 1009; i++) {
    F[i] = F[i - 1] + F[i - 2];
    F[i] %= 1001;
  }
}
int main() {
  int n, d;
  fib();
  while (cin >> n >> d) {
    UnionFind U(n);
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (abs(F[i] - F[j]) < d) {
          U.unit(i, j);
        }
      }
    }
    set<int> s;
    for (int i = 0; i < n; i++) s.insert(U.find(i));
    cout << s.size() << endl;
  }
}
