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<algorithm>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<queue>
#include<cstdio>
#include<climits>
#include<cmath>
#include<cstring>
#include<string>
#include<sstream>
#include<numeric>
#include<cassert>

#define f first
#define s second
#define mp make_pair

#define REP(i,n) for(int i=0; i<(int)(n); i++)
#define FOR(i,c) for(__typeof((c).begin()) i=(c).begin(); i!=(c).end(); i++)
#define ALL(c) (c).begin(), (c).end()
#define IN(x,s,g) ((x) >= (s) && (x) < (g))
#define ISIN(x,y,w,h) (IN((x),0,(w)) && IN((y),0,(h)))
#define print(x) printf("%d\n",x);
#define MAX (10000000)

using namespace std;

typedef unsigned int uint;
typedef long long ll;

const int _dx[] = {0,1,0,-1};
const int _dy[] = {-1,0,1,0};

int getInt(){
  int ret = 0,c;
  c = getchar();
  while(!isdigit(c)) c = getchar();
  while(isdigit(c)){
    ret *= 10;
    ret += c - '0';
    c = getchar();
  }
  return ret;
}

const int N = 1001;
int label[N+1];

int ds(int i){
  if(label[i] == i) return i;
  return label[i] = ds(label[i]);
}

int main(){
  int v,d;
  int fib[N+1];
  fib[0] = 1; fib[1] = 2;
  for(int i=2; i<=N; i++)
    fib[i] = (fib[i-1] + fib[i-2]) % N;

  while(~scanf("%d%d",&v,&d)){
    REP(i,v+1) label[i] = i;
    for(int i=1; i<=v; i++){
      for(int j=i+1; j<=v; j++){
        if(abs(fib[i] - fib[j]) < d){
          int l1 = ds(i);
          int l2 = ds(j);
          label[l1] = label[l2] = min(l1,l2);
        }
      }
    }
    int ans = 0;
    for(int i=1; i<=v; i++)
      if(i == ds(i)) ans++;
    print(ans);
  }
  return 0;
}