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>
#include<stdio.h>
using namespace std;
#define FOR(i,k,n) for(int i = (k); i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) begin(a),end(a)
#define MS(m,v) memset(m,v,sizeof(m))
#define D10  fixed<<setprecision(10)
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int, int> P;
typedef long long ll;
const ll INF = 114514810;
const ll MOD = 1000000007;
const double EPS = 1e-8;
const double PI = acos(-1.0);
struct edge
{
	int from, to, cost;
	bool operator < (const edge& e) const { return cost < e.cost; }
	bool operator >(const edge& e) const { return cost > e.cost; }
};
int dx[] = { -1, 0, 0, 1 }; int dy[] = { 0, -1, 1, 0 };
bool valid(int x, int y, int h, int w) { return (x >= 0 && y >= 0 && x < h&&y < w); }
///*************************************************************************************///
///*************************************************************************************///
///*************************************************************************************///

int fib[1001];

int main()
{
	fib[0] = 2, fib[1] = 3;
	FOR(i, 2, 1001) fib[i] = (fib[i - 1] + fib[i - 2]) % 1001;
	int V, d;
	while (cin >> V >> d)
	{
		vi v;
		REP(i, V) v.push_back(fib[i]);
		sort(ALL(v));
		int cnt = 1;
		REP(i, V-1)
		{
			if (v[i + 1] - v[i] >= d) cnt++;
		}
		cout << cnt << endl;
	}
}