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 <string>
#include <stack>
#include <cstdlib>
#include <queue>
#include <vector>
#include <set>
#include <stdio.h>
#include <cstdio>
using namespace std;
int p[1005];
int root(int a) {
	if (p[a] == a) return a;
	return (p[a] = root(p[a]));
}
void unite(int a,int b){
	p[root(a)] = root(b);
}
void same(int a ,int b){
	if (root(a) == root(b)) cout << "1" <<endl;
	else cout << "0" << endl;
}
bool rcheck(int a){
	if (root(a) == a) return 1;
	else return 0;
}
long long int F[1005];
int main() {
	int V,D;
	F[0]=1, F[1] =2;
	for (int i = 2;i< 1005;i++ ){
		F[i] = F[i-1]%1001 + F[i-2]%1001; 
		F[i] = F[i]%1001;
	}
	while ( cin >> V >> D) {
		int cou = 0;
		for (int i = 0; i < 1001; i++){
		p[i] = i;
		}
		for (int i=1;i <= V;i++) {
			for (int j=i+1;j <= V;j++ ){
				if (F[j] - F[i] < D && F[j] - F[i] > - D )
					unite(i,j);
			}

			}
		for (int i = 1;i <= V;i++){
			if (rcheck(i) == 1) {
				cou += 1;
			}
		}
		cout << cou <<  endl;
}
	}