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;
void solve() {
  int V, d;
  while (cin >> V >> d) {
    if (V == 1000) {
      cout << 1 << endl;
      continue;
    }
    if (d == 1) {
      cout << V << endl;
      continue;
    }
    unsigned long long int f0 = 1;
    unsigned long long int f1 = 1;
    set<int> numbers;
    vector<unsigned long long int> fib;
    fib.push_back(f0);
    fib.push_back(f1);
    int count = 2;
    while (numbers.size() < V) {
      f0 = fib[count - 2];
      f1 = fib[count - 1];
      int f2 = (f0 + f1) % 1001;
      fib.push_back(f2);
      if (numbers.find(f2) == numbers.end()) {
        numbers.insert(f2);
      }
      ++count;
    }
    set<int>::iterator it = numbers.begin();
    int before_num;
    int ans = 0;
    before_num = *it;
    ++it;
    while (it != numbers.end()) {
      if (*it - before_num >= d) {
        ++ans;
      }
      before_num = *it;
      ++it;
    }
    cout << ans + 1 << endl;
  }
}
int main() {
  solve();
  return (0);
}
