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;
const int MAX_N = 1000;
class UnionFind {
 public:
  int par[MAX_N];
  int rank[MAX_N];
  UnionFind(int n) {
    for (int i = 0; i < n; ++i) {
      par[i] = i;
      rank[i] = 0;
    }
  }
  int find(int x) {
    if (par[x] == x)
      return x;
    else
      return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (rank[x] < rank[y])
      par[x] = y;
    else {
      par[y] = x;
      if (rank[x] == rank[y]) rank[x]++;
    }
  }
  bool same(int x, int y) { return find(x) == find(y); }
};
int main() {
  int v, d;
  while (cin >> v >> d) {
    int f[v];
    for (int i = 0; i < v; ++i) {
      if (i == 0)
        f[i] = 1 + 1;
      else if (i == 1)
        f[i] = f[i - 1] + 1;
      else
        f[i] = (f[i - 1] + f[i - 2]) % 1001;
    }
    UnionFind u(v);
    for (int i = 0; i < v - 1; ++i)
      for (int j = i + 1; j < v; ++j)
        if (abs(f[i] - f[j]) < d) u.unite(i, j);
    int ans;
    for (int i = 0; i < v; ++i)
      if (u.find(i) == i) ans++;
    cout << ans << endl;
  }
  return 0;
}
