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 <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <complex>
#include <ctime>
#include <cstdlib>

using namespace std;

inline int to_int(string s) {int v; istringstream sin(s); sin >> v; return v;}
template<class T> inline string to_str(T x) {ostringstream sout; sout << x; return sout.str();}

typedef long long ll;

ll par[1028];
int rank[1028];

void init(int n)
{
	for(int i = 0; i < n; i++)
	{
		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;
	}
	else
	{
		if(rank[x] < rank[y])
		{
			par[x] = y;
		}
		else
		{
			par[y] = x;
			if(rank[x] == rank[y])
			{
				rank[x]++;
			}
		}
	}
}

int main()
{
	int V, d;
	ll T[1028];
	while(cin >> V >> d)
	{
		init(V);

		T[0] = 1; T[1] = 2;
		for(int i = 2; i <= V+1; i++)
		{
			T[i] = T[i-1] + T[i-2];
		}
		for(int i = 1; i <= V+1; i++)
		{
			T[i] = T[i] % 1001;
		}

		for(int i = 1; i <= V; i++)
		{
			for(int j = i+1; j <= V; j++)
			{
				if(	abs(T[j]-T[i]) < d)
				{
					unite(i, j);
				}
			}
		}

		int ans = 0;
		for(int i = 1; i <= V; i++)
		{	
			if(par[i] == i)
			{
				ans++;
			}
		}

		cout << ans << endl;
	}
}