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>
using namespace std;
const int N = 1001;
int f[1000] = {2, 3};
int par[1000];

void init()
{
	for (int i = 0; i < 1000; i++) par[i] = i;
}

int find(int x)
{
	if (par[x] == x) return x;
	return par[x] = find(par[x]);
}

void _union(int x, int y)
{
	x = find(x);
	y = find(y);
	if (x == y) return;
	par[x] = y;
}

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

int main()
{
	for (int i = 2; i < 1000; i++){
		f[i] = (f[i - 2] + f[i - 1]) % N;
	}
	
	int v, d;
	while (scanf("%d %d", &v, &d) != EOF){
		init();
		for (int i = 0; i < v; i++){
			for (int j = 1; j < v; j++){
				if (abs(f[i] - f[j]) < d){
					_union(i, j);
				}
			}
		}
		
		int res = 0;
		for (int i = 0; i < v; i++){
			res += (i == par[i]);
		}
		printf("%d\n", res);
	}
	
	return 0;
}