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>
#include <algorithm>
#include <cmath>
using namespace std;

const int C = 1010;
const int N = 1001;
const int UFSIZE = C;
int P[UFSIZE]; // 親

void init() {
  for (auto i=0; i<UFSIZE; i++) {
    P[i] = i;
  }
}

int root(int a) {
  if (a == P[a]) return a;
  return (P[a] = root(P[a]));
}

bool issame(int a, int b) {
  return root(a) == root(b);
}

void unite(int a, int b) {
  P[root(a)] = root(b);
}

bool isroot(int a) {
  return root(a) == a;
}

int main() {
  int V, d;
  int Fib[C];
  Fib[0] = 1;
  Fib[1] = 2;
  for (auto i=2; i<C; i++) {
    Fib[i] = (Fib[i-1] + Fib[i-2])%N;
  }
  while (cin >> V >> d) {
    init();
    for (auto i=1; i<=V; i++) {
      for (auto j=i+1; j<=V; j++) {
	if (abs(Fib[i]-Fib[j]) < d) {
	  unite(i, j);
	}
      }
    }
    int ans = 0;
    for (auto i=1; i<=V; i++) {
      if (isroot(i)) ans++;
    }
    cout << ans << endl;
  }
}