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;
const int INF = 1 << 29, MOD = 1e9 + 7;
int v[10010];
void init(int n) {
  for (int i = 0; i < n; i++) v[i] = i;
}
int root(int a) {
  if (v[a] == a) return a;
  return (v[a] = root(v[a]));
}
bool is_same_set(int a, int b) { return root(a) == root(b); }
void unite(int a, int b) { v[root(a)] = root(b); }
int main() {
  int V, d;
  while (cin >> V >> d) {
    init(V);
    int fib[V + 2];
    fib[0] = 2;
    fib[1] = 3;
    for (int i = 0; i < V; i++) fib[i + 2] = (fib[i + 1] + fib[i]) % 1001;
    for (int i = 0; i < V; i++)
      for (int j = 0; j < V; j++) {
        if (abs(fib[i] - fib[j]) < d) unite(j, i);
      }
    sort(v, v + V);
    int cnt = 1;
    for (int i = 0; i < V - 1; i++) {
      if (v[i] != v[i + 1]) cnt++;
    }
    cout << cnt << endl;
  }
  return 0;
}
