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[1005];
int root(int a) {
  if (p[a] == a) return a;
  return (p[a] = root(p[a]));
}
void unite(int a, int b) { p[root(a)] = root(b); }
void same(int a, int b) {
  if (root(a) == root(b))
    cout << "1" << endl;
  else
    cout << "0" << endl;
}
bool rcheck(int a) {
  if (root(a) == a)
    return 1;
  else
    return 0;
}
long long int F[1005];
int main() {
  int V, D;
  F[0] = 1, F[1] = 2;
  for (int i = 2; i < 1005; i++) {
    F[i] = F[i - 1] % 1001 + F[i - 2] % 1001;
  }
  while (cin >> V >> D) {
    int cou = 0;
    for (int i = 0; i < 1001; i++) {
      p[i] = i;
    }
    for (int i = 1; i <= V; i++) {
      for (int j = 1; j <= V; j++) {
        if (F[j] - F[i] < D && F[j] - F[i] > -D) unite(i, j);
      }
    }
    for (int i = 1; i <= V; i++) {
      if (rcheck(i) == 1) {
        cou += 1;
      }
    }
    cout << cou << endl;
  }
}
