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;
const int N = 1001;
long long int func(int n, long long int a[1002]) {
  if (n <= 0) {
    return 1;
  }
  if (n == 1) {
    return 2;
  }
  if (a[n] != 0) {
    return a[n];
  }
  return a[n] = func(n - 1, a) + func(n - 2, a);
}
int main() {
  int v, d, s;
  long long int a[1001], b[1001];
  fill(a, a + 1002, 0);
  for (int i = 0; i <= N; i++) {
    a[i] = func(i, a);
  }
  while (cin >> v >> d) {
    for (int i = 1; i <= v; i++) {
      b[i - 1] = a[i] % N;
    }
    sort(b, b + v);
    s = 1;
    for (int i = 1; i < v; i++) {
      if (b[i] - b[i - 1] >= d) {
        s++;
      }
    }
    cout << s << endl;
  }
  return 0;
}
