Problem:
Problem H: Fibonacci Sets

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?

Figure 1: There are 4 connected subsets for V = 20 and d = 100.

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.

Constraints

1 ≤ V ≤ 1000

1 ≤ d ≤ 150

Output

Output line contains one integer - the number of connected subsets - for each input line.

Sample Input

5 5
50 1
13 13

Output for the Sample Input

2
50
8

Solution in C++ (judge verdict: Accepted)

#include <iostream>
#include <set>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <vector>



#define MAX_N 10000
#define INF   100000000

using namespace std;


long long par[MAX_N];  //parent 0f node
long long rank[MAX_N]; //depth of node

long long fib[MAX_N];



//n個ので初期化、
void init(long long n) { 
    for (long long i=1; i <= n; i++){
        par[i] = i;
        rank[i] = 0;
    }
}

//代表ノードの親は自分自身
long long find(long long x) { 
    if (par[x] == x) {
        return x;
    } else {
        par[x] = find(par[x]);
        return par[x];
    }
}

void unite(long long x, long long y){
    x = find(x);
    y = find(y);
    if (x==y) { return; }
    else if (rank[x] < rank[y]) { 
        par[x] = y;
    }
    else if (rank[y] < rank[x]) { 
        par[y] = x;
    }
    else {
        par[y] = x;
        rank[x]++;
    }
}

void init_fib(){
    for( long long i = 0; i<MAX_N; i++){
        fib[i] = INF;
    }
}

long long f(long long n){
    if(n == 0 || n==-1) { return 1;} 
    else if(fib[n] != INF ) {return fib[n];}
    else { return fib[n] = (  f(n-1) + f(n-2) )%1001 ; }
}

void print_fib(){
    for ( long long i = 1; i<=10 ; i++ ) {
        cout << f(i) << " " ;
    }
    cout << endl;
}

int main()
{
    init_fib();
    //cout << f(100);
    //print_fib();

    long long V,d;
    while( cin >> V >>d ){
       if( V==0 && d==0){ break;}
       init(MAX_N);
       for( long long i=1; i<=V; i++ ){
           for( long long j=i + 1; j<=V; j++ ){
               if( abs( f(i) - f(j) )%1001 < d ){
                   //cout << "unite(" << i << "," << j << ")" << endl;
                   //cout << f(i) << "-" << f(j) << " = " <<  abs( f(i) - f(j) ) << endl;
                   unite(i,j);
               }
           }
       }

       set<long long> count;
       for( long long i=1; i<=V; i++){
           //cout << "count = " << find(i) << endl;
           count.insert( find(i) );
       }
       cout << count.size() << endl;
    }
    return 0;
}