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 <iomanip>
#include <sstream>
#include <stdio.h>
#include <list>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <math.h>
#include <utility>
#include <string>
#include <ctype.h>
#include <cstring>
#include <cstdio>
#include <sstream>
#include <functional>

using namespace std;

#define FOR(i,k,n) for(int i = (k); i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define INF 114514810
#define ll long long
//#define scanf scanf_s

int v, d, fib[1005];
int par[1005];

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

void unite(int x, int y)
{
	x = par[x];
	y = par[y];
	par[x] = y;
}


int main()
{
	fib[0] = 1; fib[1] = 2;
	REP(i, 1002)
	{
		fib[i + 2] = (fib[i + 1] + fib[i]) % 1001;
	}
	while (cin >> v >> d)
	{
		REP(i, 1005) par[i] = i;
		FOR(i, 1, v+1)FOR(j, 1, v+1)
		{
			if (abs(fib[i] - fib[j]) < d && find(i) != find(j))
			{
				unite(i, j);
			}
		}
		bool used[1005] = {};
		int cnt = 0;
		FOR(i, 1, v+1)
		{
			if (used[find(i)] == false)
			{
				cnt++;
				used[find(i)] = true;
			}
		}
		cout << cnt << endl;
	}
	return 0;
}