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 fib[1010];
int d, V;
int parent[1010];
void makefib() {
  fib[0] = 2;
  fib[1] = 3;
  for (int i = 2; i < 1010; i++) fib[i] = (fib[i - 1] + fib[i - 2]) % 1001;
  return;
}
void initialize() {
  for (int i = 0; i < V; i++) {
    parent[i] = i;
  }
}
int root(int x) { return (parent[x] == x) ? x : root(parent[x]); }
bool sameset(int x, int y) { return root(x) == root(y); }
void union_set(int x, int y) { parent[root(y)] = root(x); }
int main() {
  makefib();
  while (cin >> V >> d) {
    initialize();
    for (int i = 0; i < V; i++) {
      for (int j = 0; j < V; j++) {
        if (abs(fib[i] - fib[j]) < d && !sameset(i, j)) union_set(i, j);
      }
    }
    int cnt = 0;
    for (int i = 0; i < V; i++) {
      if (root(i) == i) cnt++;
    }
    cout << cnt << endl;
  }
}
