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 <iostream>
#include <complex>
#include <sstream>
#include <string>
#include <algorithm>
#include <deque>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <vector>
#include <set>
#include <limits>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
#define REP(i, j) for(int i = 0; i < (int)(j); ++i)
#define FOR(i, j, k) for(int i = (int)(j); i < (int)(k); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define REVERSE(v) reverse((v).begin(), (v).end())
const int N = 1001;

vector<int> v, parent, depth;

int find(int a){
  if(parent[a] == a) return a;
  else return parent[a] = find(parent[a]);
}

void merge(int a, int b){
  int pa = find(a), pb = find(b);
  if(pa == pb) return ;

  if(depth[pa] > depth[pb]) swap(pa, pb);
  if(depth[pa] == depth[pb]) ++depth[pa];
  parent[pb] = pa;
}

void init_union_find(int V){
  v = vector<int>(V);
  parent = vector<int>(V);
  depth = vector<int>(V, 1);
  REP(i, V) parent[i] = i;
}

int main() {
  int V, F[N];
  F[0] = 2; F[1] = 3;
  FOR(i, 2, N) F[i] = (F[i - 1] + F[i - 2]) % N;
  while(cin >>V){
    int D; cin >>D;
    vector<int> cnt(V, 0), L(V);
    REP(i, V) L[i] = F[i];
    init_union_find(V);
    REP(i, V) REP(j, V) if(abs(L[i] - L[j]) < D) merge(i, j);
    set<int> S;
    REP(i, V) S.insert(find(i));
    cout <<S.size() <<endl;
  }
  return 0;
}