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;
vector<int> par;
void init(int);
int root(int);
void uni(int, int);
bool same(int, int);
int main() {
  int V, d;
  while (cin >> V >> d) {
    init(V);
    vector<int> f(V);
    int temp1, temp2;
    for (int i = 0; i < V; i++) {
      if (i == 0)
        f[i] = temp1 = 2;
      else if (i == 1)
        f[i] = temp2 = 3;
      else {
        f[i] = temp1 + temp2;
        temp1 = temp2;
        temp2 = f[i];
      }
    }
    sort(f.begin(), f.end());
    for (int i = 0; i < V - 1; i++)
      if (f[i + 1] - f[i] < d) uni(i, i + 1);
    int count = 1;
    for (int i = 0; i < V - 1; i++)
      if (root(i) != root(i + 1)) count++;
    cout << count << endl;
  }
  return 0;
}
void init(int n) {
  par.resize(n);
  for (int i = 0; i < n; i++) {
    par[i] = i;
  }
}
int root(int x) {
  if (par[x] == x)
    return x;
  else
    return par[x] = root(par[x]);
}
void uni(int x, int y) {
  x = root(x);
  y = root(y);
  if (x == y) return;
  par[x] = y;
}
bool same(int x, int y) { return root(x) == root(y); }
