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>
using namespace std;
int par[1000];
int Rank[1000];
void init(int n) {
  for (int i = 0; i < n; i++) {
    par[i] = i;
    Rank[i] = 0;
  }
}
int find(int x) {
  if (par[x] == x) {
    return x;
  }
  return par[x] = find(par[x]);
}
void unite(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  if (Rank[x] < Rank[y]) {
    par[x] = y;
  } else {
    par[y] = x;
    if (Rank[x] == Rank[y]) Rank[x]++;
  }
}
bool same(int x, int y) { return find(x) == find(y); }
int main() {
  int V, d;
  int fib[1000];
  fib[0] = 2;
  fib[1] = 3;
  for (int i = 2; i < 1000; i++) {
    fib[i] = fib[i - 1] + fib[i - 2];
  }
  while (cin >> V >> d) {
    init(V);
    for (int i = 0; i < V; i++) {
      for (int j = 0; j < V; j++) {
        if (abs(fib[i] - fib[j]) < d) {
          unite(i, j);
        }
      }
    }
    int count = 0;
    for (int i = 0; i < V; i++) {
      if (find(i) == i) count++;
    }
    printf("%d\n", count);
  }
}
