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 <algorithm>
#include <vector>
#include <cmath>
using namespace std;

struct UnionFind {
    vector<int> par;

    UnionFind(int n) : par(n, -1) { }

    int root(int x) {
        return par[x] < 0 ? x : par[x] = root(par[x]);
    }

    bool same(int x, int y) {
        return root(x) == root(y);
    }

    void merge(int x, int y) {
        x = root(x), y = root(y);
        if(x != y) {
            par[x] += par[y]; par[y] = x;
        }
    }

    int distinct() {
        int res = 0;
        for(int i = 0; i < (int)par.size(); i++) res += (par[i] < 0);
        return res;
    }
};

const int MOD = 1001;

int main() {
    vector<int> fib(1001);
    fib[0] = 1, fib[1] = 2;
    for(int i = 2; i <= 1000; i++) {
        fib[i] = (fib[i - 1] + fib[i - 2]) % MOD;
    }
    for(int V, D; cin >> V >> D; ) {
        UnionFind uf(V);
        for(int i = 1; i <= V; i++) {
            for(int j = 1; j <= V; j++) {
                if(abs(fib[i] - fib[j]) < D) uf.merge(i - 1, j - 1);
            }
        }
        int res = uf.distinct();
        cout << res << endl;
    }
}