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<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
using ld = long double;
template<class t>
using table = vector<vector<t>>;
const ld eps = 1e-9;

//// < "d:\d_download\visual studio 2015\projects\programing_contest_c++\debug\a.txt" > "d:\d_download\visual studio 2015\projects\programing_contest_c++\debug\b.txt"
  

int main() {
	int V, D;
	int aa = 1001;
	while (cin >> V >> D) {
		vector<int>fib(V+1);
		fib[0] = 1%aa;
		fib[1] = 2%aa;
		for (int i = 2; i <= V; ++i) {
			fib[i] = (fib[i - 1] + fib[i - 2])%aa;
		}
		vector<int>nums;
		for (int i = 1; i <= V; ++i) {
			nums.emplace_back(fib[i]);
		}
		sort(nums.begin(), nums.end());
		nums.erase(unique(nums.begin(), nums.end()), nums.end());
		int from = -10000;
		int ans = 0;
		for (int i = 0; i < nums.size(); ++i) {
			if (from + D <= nums[i]) {
				ans++;


			}
			from = nums[i];
		}
		cout << ans << endl;
	}
	return 0;
}