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<algorithm>
#include<bitset>
#include<cassert>
#include<cfloat>
#include<climits>
#include<cmath>
#include<deque>
#include<functional>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

using lint = long long;
using P = pair<int, int>;
using LLP = pair<long long, long long>;

#define REP(i, x, n) for(int i = (x), i##_len = (int)(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = (int)(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = (int)(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = (int)(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = (int)(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())
#define REVERSE(x) reverse((x).begin(), (x).end())
#define TWINS(x) cout << ((x) ? "Yay!" : ":(") << '\n'
#define endl '\n'

constexpr int IINF = (1 << 30) - 1;
constexpr long long LLINF = 1LL << 61;
constexpr double EPS = 1e-10;

constexpr int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
constexpr int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};

class UnionFind{
    vector<int> par;
    int groupCount;

public:
    UnionFind(int n = 0){
        init(n);
    }

    void init(int n = 0){
        par.resize(n);
        fill(par.begin(), par.end(), -1);

        groupCount = n;

        return;
    }

    int root(int x){
        if(par[x] < 0){
            return x;
        }
        return par[x] = root(par[x]);
    }

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

    int size(int x){
        return -par[root(x)];
    }

    bool unite(int x, int y){
        x = root(x);
        y = root(y);

        if(x == y){
            return false;
        }

        if(par[y] < par[x]){
            swap(x, y);
        }

        par[x] += par[y];
        par[y] = x;
        --groupCount;

        return true;
    }

    int size(void){
        return groupCount;
    }
};

template<typename T>
bool chmax(T& a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool chmin(T& a, T b){
    if(b < a){
        a = b;
        return true;
    }
    return false;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    constexpr int n = 1001;

    int fib[n + 1];
    fib[0] = 1;
    fib[1] = 2;
    REP(i, 2, n + 1){
        fib[i] = (fib[i - 1] + fib[i - 2]) % n;
    }

    int v, d;
    while(cin >> v >> d){
        UnionFind uf(v + 1);
        for(int i = 1 ; i <= v ; ++i){
            for(int j = 1 ; j <= v ; ++j){
                if(abs(fib[i] - fib[j]) < d){
                    uf.unite(i, j);
                }
            }
        }
        cout << uf.size() - 1 << endl;
    }

    cout << flush;

    return 0;
}
