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?

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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 const N = 1001;
unsigned long long int fibs[N + 1] = {1};
unsigned long long int f(int i) { return (i <= -1) ? 1 : fibs[i]; }
int F[N + 1];
struct S {
  vector<unsigned long long int> cs;
};
int main() {
  for (int i = 1; i <= N; ++i) {
    fibs[i] = f(i - 1) + f(i - 2);
    F[i] = f(i) % N;
  }
  int V, d;
  while (cin >> V >> d) {
    map<unsigned long long, S> tb;
    int c = 1, sum = 0;
    for (int i = 1; i <= V; ++i) {
      for (int j = 1; j <= V; ++j) {
        if (abs(F[i] - F[j]) < d) {
          int lm = i < j ? i : j;
          int ln = i < j ? j : i;
          tb[lm].cs.push_back(ln);
        }
      }
    }
    while (tb.size() != 0) {
      queue<unsigned long long> q;
      q.push(tb.begin()->first);
      while (!q.empty()) {
        unsigned long long k = q.front();
        q.pop();
        S const& s = tb[k];
        for (int ii = 0; ii < s.cs.size(); ++ii) {
          q.push(s.cs[ii]);
        }
        tb.erase(k);
      }
      ++sum;
    }
    cout << sum << endl;
    ;
  }
}
