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 P[10010];
int F[1001];
int FF[1001];
int v;
int d;
void init(int N) {
  for (int i = 0; i < N; ++i) P[i] = i;
}
int root(int a) {
  if (P[a] == a) return a;
  return (P[a] = root(P[a]));
}
bool is_same_set(int a, int b) { return root(a) == root(b); }
void unite(int a, int b) { P[root(a)] = root(b); }
int main() {
  v = 0;
  d = 0;
  F[0] = 1;
  F[1] = 2;
  for (int i = 2; i < 1001; ++i) {
    F[i] = F[i - 1] + F[i - 2];
  }
  for (int i = 0; i < 1001; ++i) {
    FF[i] = F[i] % 1001;
  }
  int count;
  while (cin >> v >> d) {
    count = 0;
    init(1001);
    for (int i = 1; i <= v; ++i) {
      for (int j = i + 1; j <= v; ++j) {
        if (abs(FF[i] - FF[j]) < d) {
          if (!is_same_set(i, j)) {
            unite(i, j);
          }
        }
      }
    }
    for (int i = 0; i < v; ++i) {
      if (root(i) == i) {
        ++count;
      }
    }
    cout << count << endl;
  }
}
