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;
typedef struct UnionFindTree{
    vector<int> par;
    
    UnionFindTree(int n){
        par.resize(n, -1);
    }

    int find(int x){
        if(par[x] < 0) return x;
	return par[x] = find(par[x]);
    }

    bool unite(int x, int y){
        x = find(x);
	y = find(y);
	if(x == y) return false;
	if(par[x] > par[y]) swap(x, y);
	par[x] += par[y];
	par[y] = x;
	return true;
    }
}UF;

int main(){
  int v, d;
  int fib[1010];
  int a, b;
  a = b = 1;
  int mod = 1001;
  for(int i=0; i<1010; ++i){
    int c = (a + b) % mod;
    fib[i] = c;
    a = b;
    b = c;
  }
  while(scanf("%d%d", &v, &d) != EOF){
    int ans = v;
    UF uf(v);
    for(int i=0; i<v; ++i){
      for(int j=i+1; j<v; ++j){
	    if(abs(fib[i] - fib[j]) < d) ans -= uf.unite(i, j);
      }
    }
    cout << ans << endl;
  }
}
