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;
vector<int> fibo;
void makeFiboModN(int V) {
  fibo.resize(V + 1);
  fibo[0] = 1;
  fibo[1] = 2;
  for (int i = 2; i <= V; ++i) {
    fibo[i] = (fibo[i - 1] + fibo[i - 2]) % 1001;
  }
}
int main() {
  while (true) {
    vector<vector<int> > vs;
    int ans = 0;
    int V, d;
    cin >> V >> d;
    if (cin.eof()) break;
    makeFiboModN(V);
    vector<int> tv;
    tv.push_back(fibo[1]);
    vs.push_back(tv);
    for (int i = 2; i <= V; ++i) {
      bool bConnected = false;
      for (int j = 0; j < vs.size() && !bConnected; ++j) {
        for (int k = 0; k < vs[j].size(); ++k) {
          if (abs(vs[j][k] - fibo[i]) < d) {
            vs[j].push_back(fibo[i]);
            bConnected = true;
            break;
          }
        }
      }
      if (!bConnected) {
        tv.clear();
        tv.push_back(fibo[i]);
        vs.push_back(tv);
      }
    }
    cout << vs.size() << endl;
  }
  return 0;
}
