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 <cstdlib>
#include <cmath>

using namespace std;

long long  F[1001];

int P[10010];
void init(int N){
    for(int i=0;i<=N;i++)
        P[i] = i;
}

int root(int a){
    if(P[a]==a)
        return a;
    return(P[a] = root(P[a]));
}

bool is_same_set(int a, int b){
    return root(a) == root(b);
}

void unit(int a, int b){
    P[root(a)] = root(b);
}

int main(){
    F[0] = 1; F[1]=2;
    for(int i=2;i<=1000;i++){
        F[i] = F[i-1]+F[i-2];
        F[i] %= 1001;
    }
    int V,d;
    while(cin >> V >> d){
        int count=0;
        init(V);
        for(int i=1;i<=V;i++){
            for(int j=i+1; j<=V;j++){
                if(abs(F[j]-F[i])<d){
                    unit(i,j);
                }
            }
        }
        for(int i=1;i<=V;i++){
            if(root(i)==i)
                count++;
        }
        cout << count << endl;
    }
}