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 <vector>
#include <array>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <tuple>
#include <bitset>
#include <memory>
#include <cmath>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <numeric>
#include <climits>
#include <cfloat>
#include <cassert>
int main()
{
	constexpr int n = 1001;
	int v, d;
	while (std::cin >> v >> d) {
		std::vector<bool> exist(n, false);
		int prev = 1, current = 1;
		for (auto i = 0; i < v; ++i) {
			current = (current + prev) % n;
			prev = (current - prev + n) % n;
			exist[current] = true;
		}
		std::queue<int> queue;
		int count = 0;
		for (auto i = 0; i < n; ++i) if (exist[i]) {
			while (!queue.empty() && i - queue.front() >= d) queue.pop();
			if (queue.empty()) ++count;
			queue.push(i);
		}
		std::cout << count << '\n';
	}
}
