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>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT(n) (1LL<<(n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

typedef long long LL;
template<typename T> using V = std::vector<T>;
template<typename T> using VV = std::vector<std::vector<T>>;
template<typename T> using VVV = std::vector<std::vector<std::vector<T>>>;

template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline void print_vec(const std::vector<T> &v) { for (int i = 0; i < v.size(); ++i) {  if (i != 0) {std::cout << " ";} std::cout << v[i];} std::cout << "\n"; }
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }

const int INF = 1L << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy4 = { 0, 1, 0, -1 }, dx4 = { 1, 0, -1, 0 };    // 4近傍（右, 下, 左, 上）
const std::vector<int> dy8 = { 0, -1, 0, 1, 1, -1, -1, 1 }, dx8 = { 1, 0, -1, 0, 1, 1, -1, -1 };

using namespace std;

void dfs(int i, const vector<vector<int>> &g, vector<bool> &used) {
    used[i] = true;
    FOE(j, g[i]) {
        if (not used[j]) {
            dfs(j, g, used);
        }
    }
}

int main() {
    cin.tie(0);
	ios::sync_with_stdio(false);

	vector<int> fib(1100);
	fib[0] = 1;
	fib[1] = 2;
	FOR(i, 2, fib.size()) {
	    fib[i] = (fib[i - 1] + fib[i - 2]) % 1001;
	}

	int V, D;
    while (cin >> V >> D) {

        vector<vector<int>> g(V + 1);
        FOR(i, 1, V + 1) {
            FOR(j, 1, V + 1) {
                if (abs(fib[i] - fib[j]) < D) {
                    g[i].emplace_back(j);
                    g[j].emplace_back(i);
                }
            }
        }

        int ans = 0;
        vector<bool> used(V, false);
        FOR(i, 1, V + 1) {
            if (used[i]) {
                continue;
            }
            ans++;
            dfs(i, g, used);
        }
        print(ans);

    }

	return 0;
}
