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
#define _USE_MATH_DEFINES
#include <iostream>
#include <memory>
#include <memory.h>
#include <fstream>
#include <cmath>
#include <math.h>
#include <numeric>
#include <vector>
#include <stack>
#include <string>
#include <queue>
#include <sstream>
#include <cstdlib>
#include <cassert>
#include <cstdio>
#include <cstring>
#include <map>
#include <iomanip>
#include <list>
#include <cctype>
#include <algorithm>
#include <complex>
using namespace std;

typedef complex<double> xy_t;
typedef pair<xy_t, xy_t> line;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<double, double> Pd;
typedef pair<int, P> PP;
typedef pair<int, PP> PPP;
const ll INF = 1LL << 50;
const double EPS = 1E-10;

#define rep(i, n) for(int i = 0; i < n; i++)

int v, d;
const int mod = 1001;
bool nums[1001];
bool used[1001];

void dfs(int p){
	used[p] = true;
	//cout <<p << endl;
	for(int i = p - d + 1; i < p + d; i++){
		if(i < 0 || i >= 1001) continue;
		if(!used[i] && nums[i]){
			dfs(i);
		}
	}
}

int main(){
	
	while(cin >> v >> d){
		memset(nums, false, sizeof(nums));
		memset(used, false, sizeof(used));
		int f0 = 1;
		int f1 = 2;
		rep(i, v){
			nums[f1] = true;
			int t = f1;
			f1 = (f0 + f1) % mod;
			f0 = t;
		}
		int cnt = 0;
		rep(i, 1001){
			if(nums[i] && !used[i]){
				cnt++;
				dfs(i);
			}
		}
		cout << cnt << endl;
	}
	return 0;
}