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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

struct UnionFind {
#define rep(i,n) for (int i=0; i < int(n); i++)
  vector<int> par;
  vector<int> sizes;

  UnionFind(int n) : par(n), sizes(n, 1) {
    rep(i,n) par[i] = i;
  }

  int find(int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
  }

  void unite(int x, int y) {
    x = find(x);
    y = find(y);

    if (x == y) return;

    if (sizes[x] < sizes[y]) swap(x, y);

    par[y] = x;
    sizes[x] += sizes[y];
  }

  bool same(int x, int y) {
    return find(x) == find(y);
  }

  int size(int x) {
    return sizes[find(x)];
  }

  int count() {
    vector<int> p = par;
    sort(p.begin(), p.end());
    return unique(p.begin(), p.end()) - p.begin();
  }
#undef rep
};


const int N = 1001;

int main()
{
    for (;;) {
        int v, d;
        if (scanf("%d %d", &v, &d) != 2)
            break;

        vector<int> f(v);
        f[0] = 2;
        f[1] = 3;
        for (int i = 2; i < v; ++i)
            f[i] = (f[i - 2] + f[i - 1]) % N;
        sort(f.begin(), f.end());
        f.erase(unique(f.begin(), f.end()), f.end());

        UnionFind uf(f.size());
        for (int i = 1; i < f.size(); ++i)
            if (f[i] - f[i - 1] < d)
                uf.unite(i - 1, i);

        printf("%d\n", uf.count());
    }
    return 0;
}