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 parent[1001];
  UnionFind() { memset(parent, -1, sizeof(parent)); }
  bool Unite(int x, int y) {
    x = Root(x);
    y = Root(y);
    if (x == y) return false;
    if (parent[y] < parent[x]) {
      swap(x, y);
    }
    parent[x] += parent[y];
    parent[y] = x;
    return true;
  }
  bool Find(int x, int y) { return Root(x) == Root(y); }
  int Root(int x) { return parent[x] < 0 ? x : parent[x] = Root(parent[x]); }
};
int V, D, N = 1001;
int fib[1001];
void Init() {
  fib[0] = 2;
  fib[1] = 3;
  for (int i = 2; i < N + 1; i++) fib[i] = (fib[i - 1] + fib[i - 2]) % N;
}
void Solve() {
  UnionFind uf;
  for (int i = 0; i < V; i++) {
    for (int j = i + 1; j < V; j++) {
      if (abs(fib[i] - fib[j]) < D) {
        uf.Unite(i, j);
      }
    }
  }
  int res = 0;
  for (int i = 0; i < V; i++)
    if (uf.parent[i] != 0) res++;
  cout << res << endl;
}
int main() {
  Init();
  while (cin >> V >> D) {
    Solve();
  }
}
