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 P[100000];
int fibo[1000];
void init(int n) {
  for (int i = 0; i < n; ++i) {
    P[i] = i;
  }
}
int root(int n) {
  if (P[n] == n) {
    return n;
  }
  return root(P[n]);
}
void unite(int a, int b) { P[root(a)] = root(b); }
bool is_in_the_same_tree(int a, int b) { return root(a) == root(b); }
int num_of_trees(int n) {
  int i, roothere, sum = 0;
  vector<int> roots;
  for (i = 0; i < n; ++i) {
    roothere = root(i);
    if (find(roots.begin(), roots.end(), roothere) == roots.end()) {
      roots.push_back(roothere);
    }
  }
  return roots.size();
}
int fibonacci(int n) {
  if (n == 0) {
    fibo[n] = 2;
    return 2;
  } else if (n == 1) {
    fibo[n] = 3;
    return 3;
  } else {
    if (fibo[n] != 0) {
      return fibo[n];
    } else if (fibo[n - 1] != 0 && fibo[n - 2] != 0) {
      fibo[n] = (fibo[n - 1] + fibo[n - 2]) % 1001;
      return fibo[n];
    } else {
      int result = (fibonacci(n - 1) + fibonacci(n - 2)) % 1001;
      fibo[n] = result;
      return result;
    }
  }
}
int main() {
  int V, d, i;
  for (i = 0; i < 1000; ++i) {
    fibo[i] = 0;
  }
  while (cin >> V >> d) {
    fibonacci(V);
    init(V);
    if (V == 1) {
      cout << 1 << endl;
    } else {
      for (i = 0; i < V - 1; ++i) {
        if (abs(fibo[i + 1] - fibo[i]) < d) {
          unite(i, i + 1);
        }
      }
      cout << num_of_trees(V) << endl;
    }
  }
}
