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[1001];
int F[1001];
int V, D;
int abs(int a) {
  if (a < 0) {
    return a * (-1);
  } else {
    return a;
  }
}
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() {
  F[0] = 1, F[1] = 2;
  for (int i = 2; i < 1001; i++) {
    F[i] = F[i - 1] + F[i - 2];
  }
  while (cin >> V >> D) {
    init(V);
    for (int i = 1; i <= V; i++) {
      for (int j = i + 1; j <= V; j++) {
        if (abs((F[i] % 1001) - (F[j] % 1001)) < D) {
          unite(i, j);
        }
      }
    }
    int ans = 0;
    for (int i = 1; i <= V; i++) {
      if (root(i) == i) {
        ans += 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
