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;
int V, d;
const int N = 1001;
int F[1024];
int abs(int x) { return (x > 0 ? x : -x); }
const int VMAX = 1024;
int par[VMAX], Rank[VMAX];
int find(int x) {
  if (par[x] == x) {
    return x;
  } else {
    return par[x] = find(par[x]);
  }
}
void unite(int x, int y) {
  x = find(x), y = find(y);
  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) { return find(x) == find(y); }
int count_sets(int begin, int end) {
  set<int> s;
  for (int i = begin; i <= end; i++) {
    s.insert(par[i]);
  }
  return s.size();
}
int solve() {
  for (int i = 1; i <= V; i++)
    for (int j = 1; j <= V; j++) {
      if (abs(F[i] - F[j]) < d) unite(i, j);
    }
  return count_sets(1, V);
}
void init() {
  for (int i = 0; i < (int)(V); i++) {
    par[i] = i;
    Rank[i] = 0;
  }
}
int main() {
  F[0] = 1, F[1] = 2;
  for (int i = 2; i < 1024; i++) {
    F[i] = (F[i - 1] + F[i - 2]) % N;
  }
  while (cin >> V >> d) {
    init();
    cout << solve() << endl;
  }
}
