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?

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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 v, d, fib[1005];
int par[1005];
int find(int x) {
  if (x == par[x])
    return x;
  else
    return par[x] = find(par[x]);
}
void unite(int x, int y) {
  x = par[x];
  y = par[y];
  par[x] = y;
}
int main() {
  fib[0] = 1;
  fib[1] = 2;
  for (int i = (0); i < (1002); i++) {
    fib[i + 2] = (fib[i + 1] + fib[i]) % 1001;
  }
  while (cin >> v >> d) {
    for (int i = (0); i < (1005); i++) par[i] = i;
    for (int i = (0); i < (v); i++)
      for (int j = (0); j < (v); j++) {
        if (abs(fib[i] - fib[j]) < d && find(i) != find(j)) {
          unite(i, j);
        }
      }
    bool used[1005] = {};
    int cnt = 0;
    for (int i = (1); i < (v + 1); i++) {
      if (used[find(i)] == false) {
        cnt++;
        used[find(i)] = true;
      }
    }
    cout << cnt << endl;
  }
  return 0;
}
