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 <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>

using namespace std;

#define rep(i,j) REP((i), 0, (j))
#define REP(i,j,k) for(int i=(j);(i)<(k);++i)
#define BW(a,x,b) ((a)<=(x)&&(x)<=(b))
#define F first
#define S second
#define INF 1 << 30
#define N 1024

typedef pair<int, int> pi;
typedef pair<int, pi> pii;
typedef vector<int> vi;
typedef queue<int> qi;
typedef long long ll;

int par[N], rank[N];

void init(int n){
  rep(i, n){
    par[i] = i;
    rank[i] = 0;
  }
}

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 V, d;
int f[N];
int main(){
  f[0] = 1; f[1] = 2;
  REP(i, 2, N) f[i] = f[i-1] + f[i-2];

  while(~scanf("%d%d", &V, &d)){
    init(V+1);
    REP(i, 1, V+1){
      REP(j, i+1, V+1){
	if((abs(f[j]-f[i])%1001) < d) unite(i, j);
      }
    }
  int res = 0;
  REP(i, 1, V+1) if(par[i] ==  i) res++;
  printf("%d\n", res);
  }
  return 0;
}