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 <vector>
#include <string>
#include <map>
#include <algorithm>
#include <cstdio>
#include <set>

#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;

int uf[1001];
void inituf() {
  for (int i = 0; i < 1001; ++i) {
    uf[i] = -1;
  }
}

int root(int i) {
  if(uf[i] < 0) {
    return i;
  } else {
    return uf[i] = root(uf[i]);
  }
}

void join(int i, int j) {
  int ri = root(i);
  int rj = root(j);
  if (uf[ri] > uf[rj]) {
    uf[ri] = rj;
  } else {
    if (uf[ri] == uf[rj])
      --uf[ri];
    uf[rj] = ri;
  }
}

int main() {
  int v,d;
  while(scanf("%d%d",&v,&d) != EOF) {
    vector<int> a;
    int fibo[v+3];
    fibo[0] = 1;
    fibo[1] = 2;
    a.push_back(2);
    for (int i = 2;i <= v; ++i) {
      fibo[i] = fibo[i-1] + fibo[i-2];
      fibo[i] %= 1001;
      a.push_back(fibo[i]);
    }
    sort(begin(a),end(a));
    auto itr = unique(begin(a),end(a));
    a.resize(itr - begin(a));
    inituf();
    for (int i = 1; i < a.size(); ++i) {
      if (a[i] - a[i-1] < d) {
        join(i,i-1);
      }
    }
    set<int> rts;
    for (int i = 0; i < a.size(); ++i) {
      rts.insert(root(i));
    }
    cout << rts.size() << endl;
  }
  return 0;
}