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;
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;
int fib_table[2000000];
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);
}
void tree_init(int n, int* par, int* rank) {
  for (int i = 0; i < n; i++) {
    par[i] = i;
    rank[i] = 0;
  }
}
int tree_find(int x, int* par, int* rank) {
  if (par[x] == x) {
    return x;
  } else {
    return par[x] = tree_find(par[x], par, rank);
  }
}
void tree_unite(int x, int y, int* par, int* rank) {
  x = tree_find(x, par, rank);
  y = tree_find(y, par, rank);
  if (x == y) return;
  if (rank[x] < rank[y]) {
    par[x] = y;
  } else {
    par[y] = x;
    if (rank[x] == rank[y]) rank[x]++;
  }
}
bool same(int x, int y, int* par, int* rank) {
  return tree_find(x, par, rank) == tree_find(y, par, rank);
}
int main() {
  memset(fib_table, -1, sizeof(fib_table));
  int V, d;
  while (~scanf("%d %d", &V, &d)) {
    int rank[N];
    int par[N];
    tree_init(N, par, rank);
    vector<int> fibs;
    for (int i = 1; i <= V; i++) {
      fibs.push_back(fib(i));
      for (int j = i + 1; j <= V; j++) {
        if (abs(fib(i) - fib(j)) < d) {
          tree_unite(fib(i), fib(j), par, rank);
        }
      }
    }
    set<int> s;
    for (int i = 0; i < fibs.size(); i++) {
      s.insert(par[fibs[i]]);
    }
    cout << s.size() << endl;
  }
}
