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 {
  vector<int> par;
  vector<int> sizes;
  UnionFind(int n) : par(n), sizes(n, 1) {
    for (int i = 0; i < int(n); i++) par[i] = i;
  }
  int find(int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (sizes[x] < sizes[y]) swap(x, y);
    par[y] = x;
    sizes[x] += sizes[y];
  }
  bool same(int x, int y) { return find(x) == find(y); }
  int size(int x) { return sizes[find(x)]; }
  int count() {
    vector<int> p = par;
    sort(p.begin(), p.end());
    return unique(p.begin(), p.end()) - p.begin();
  }
};
int fibo(int n) {
  if (n <= 0) {
    return 1;
  }
  return fibo(n - 1) + fibo(n - 2);
}
int main() {
  int v, d;
  scanf("%d %d", &v, &d);
  int a[v];
  a[0] = 2;
  a[1] = 3;
  for (int i = 2; i < v; i++) {
    a[i] = (a[i - 1] + a[i - 2]) % 1001;
  }
  UnionFind uf(v);
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      int m = a[i] - a[j];
      if ((m > 0 && m < d) || (m < 0 && -m < d) || m == 0) {
        uf.unite(i, j);
      }
    }
  }
  cout << uf.count() << endl;
}
