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;
int par[1010], h[1010];
void init(int n) {
  for (int i = 0; i < n; i++) {
    par[i] = i;
    h[i] = 0;
  }
}
int find(int x) {
  if (par[x] == x) return x;
  return par[x] = find(par[x]);
}
bool same(int a, int b) { return find(a) == find(b); }
void unite(int a, int b) {
  a = find(a);
  b = find(b);
  if (same(a, b)) return;
  if (h[a] < h[b])
    par[a] = b;
  else {
    par[b] = a;
    if (h[a] == h[b]) h[a]++;
  }
}
const int N = 1001;
int main() {
  int v, d, f[1010];
  f[0] = f[1] = 1;
  for (int i = 0; i < N; i++) f[i + 2] = (f[i] + f[i + 1]) % N;
  while (cin >> v >> d) {
    init(v);
    for (int i = 0; i < v; i++) {
      for (int j = i + 1; j < v; j++) {
        if (abs(f[i + 2] - f[j + 2]) < d) unite(i, j);
      }
    }
    set<int> s;
    for (int i = 0; i < v; i++) s.insert(find(i));
    cout << s.size() << endl;
  }
}
