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>
int fibo[10001];
int p[10001];
int getfibo(int n) {
  if (n < 1) return 1;
  if (fibo[n]) return fibo[n];
  return fibo[n] = (getfibo(n - 1) + getfibo(n - 2)) % 1001;
}
int root(int a) { return p[a] < 0 ? a : (p[a] = root(p[a])); }
int find(int a, int b) { return root(a) == root(b); }
void merge(int a, int b) { p[root(a)] = root(b); }
int main(void) {
  fibo[0] = 1;
  fibo[1] = 2;
  int a, d, i, j;
  while (scanf("%d%d", &a, &d) == 2) {
    memset(p, -1, sizeof(p));
    for (i = 1; i <= a; i++) {
      for (j = 1; j <= a; j++) {
        if (abs(getfibo(i) - getfibo(j)) < d) {
          if (!find(i, j)) merge(i, j);
        }
      }
    }
    int b[10001] = {0}, res, c = 0;
    for (i = 1; i <= a; i++) {
      res = root(i);
      if (!b[res]) {
        b[res] = 1;
        c++;
      }
    }
    printf("%d\n", c);
  }
  return 0;
}
