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 main() {
  int V, d;
  vector<int> ans;
  while (cin >> V) {
    ans.clear();
    cin >> d;
    int rank[V + 1], parent[V + 1], label[V + 1];
    label[0] = 1;
    label[1] = 2;
    for (int i = 0; i < V + 1; i++) {
      rank[i] = 1;
      parent[i] = i;
      if (i >= 2) label[i] = (label[i - 1] + label[i - 2]) % 1001;
    }
    for (int i = 1; i < V + 1; i++) {
      for (int j = 1; j < V + 1; j++) {
        if (abs(label[i] - label[j]) < d) {
          int now1 = i, now2 = j;
          while (parent[now1] != now1) {
            now1 = parent[now1];
          }
          while (parent[now2] != now2) {
            now2 = parent[now2];
          }
          if (rank[now1] > rank[now2]) {
            parent[now2] = now1;
            ans.push_back(now1);
          } else if (rank[now1] < rank[now2]) {
            parent[now1] = now2;
            ans.push_back(now2);
          } else {
            parent[now2] = now1;
            ans.push_back(now1);
            rank[now1]++;
          }
        }
      }
    }
    sort(ans.begin(), ans.end());
    ans.erase(unique(ans.begin(), ans.end()), ans.end());
    int trueans = ans.size();
    for (int i = 0; i < ans.size(); i++) {
      for (int j = 0; j < ans.size(); j++) {
        if (i != j) {
          if (abs(label[ans[i]] - label[ans[j]]) < d) {
            trueans--;
          }
        }
      }
    }
    cout << trueans << endl;
  }
  return 0;
}
