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<string>
#include<algorithm>
#include<cmath>
#include<map>
#include<vector>
#include<deque>
using namespace std;

int main() {

    int v, d;
    while(cin >> v >> d && (v || d)) {
        deque<int> fib (v);
        fib[0] = 2;
        fib[1] = 3;
        for(int i = 2; i < v; i++) {
            fib[i] = (fib[i-1] + fib[i-2]) % 1001;
        }
        sort(fib.begin(), fib.end());

        int sets = 1;
        int now = fib.front();
        fib.pop_front();
        while(!fib.empty()) {
            if(fib.front() - now >= d) {
                sets += 1;
            }
            now = fib.front();
            fib.pop_front();
        }

        cout << sets << endl;

    }
     
    return 0;
}