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;
static const double EPS = 1e-8;
const int tx[] = {0, 1, 0, -1};
const int ty[] = {-1, 0, 1, 0};
const static long long N = 1001;
short fib_table[10000000];
int fib(int n) {
  if (n == -1 || n == 0)
    return 1;
  else if (fib_table[n] != -1)
    return fib_table[n];
  else
    return (fib_table[n] = ((fib(n - 1) % N) + (fib(n - 2) % N)) % N);
}
int main() {
  memset(fib_table, -1, sizeof(fib_table));
  int V, d;
  while (~scanf("%d %d", &V, &d)) {
    int rank[N];
    int par[N];
    vector<int> fibs;
    for (int i = 1; i <= V; i++) {
      fibs.push_back(fib(i));
    }
    sort(fibs.begin(), fibs.end());
    int res = 0;
    for (int i = 1; i < fibs.size(); i++) {
      if (fibs[i] - fibs[i - 1] >= d) res++;
    }
    cout << res << endl;
  }
}
