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 <iostream>
using namespace std;
int F[1001];
int P[10010];
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]) % 1001;
 }
 int V, D;
 while (cin >> V >> D) {
  init(V+1);
  for (int i=1; i<=V; ++i) {
   for (int j=i+1; j<=V; ++j) {
    if(F[j]-F[i]<D && -D<F[j]-F[i]) { unite(i, j); }
   }
  }
  int ans = 0;
  for (int i=1; i<=V; ++i) {
   if (root(i) == i) { ans++; }
  }
  cout << ans << endl;
 }
}
