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 <sstream>
#include <fstream>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <numeric>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cmath>
using namespace std;

typedef long long INT;
const INT mod = 1001;

int F[1001];

struct UnionFind {
  vector<int> data;
  UnionFind(int size) : data(size, -1) { }
  bool unionSet(int x, int y) {
    x = root(x); y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y]; data[y] = x;
    }
    return x != y;
  }
  bool findSet(int x, int y) {
    return root(x) == root(y);
  }
  int root(int x) {
    return data[x] < 0 ? x : data[x] = root(data[x]);
  }
  int size(int x) {
    return -data[root(x)];
  }
};

int main() {
    F[0] = 2;
    F[1] = 3;
    for ( int i = 2; i < 1001; i++ ) {
        F[i] = ( F[i-1] + F[i-2] ) % mod;
    }

    int V, d;
    while ( cin >> V >> d ) {
        UnionFind uf(V);
        V = V % mod;
        for ( int i = 0; i < V; i++ ) {
            for ( int j = i+1; j < V; j++ ) {
                if ( abs( F[i] - F[j] ) < d ) {
                    uf.unionSet(i,j);
                }
            }
        }
        int answer = 0;
        for ( int i = 0; i < V; i++ ) {
            if ( uf.root(i) == i ) answer++;
        }
        cout << answer << endl;
    }

    return 0;
}