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?

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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 <bits/stdc++.h>
using namespace std;
const int N = 1001;
int V, d;
long long f[1001];
int F[1002];
int G[1001][1001];
bool used[1001];
void dfs(int s) {
  used[s] = 1;
  cout << F[s] << " ";
  for (int i = 0; i < V; i++) {
    if (!used[i] && G[s][i]) dfs(i);
  }
}
int main() {
  while (cin >> V >> d) {
    fill(G[0], G[1001], 0);
    f[0] = F[0] = 2;
    f[1] = F[1] = 3;
    for (int i = 2; i < V; i++) {
      f[i] = f[i - 1] + f[i - 2];
      F[i] = f[i] % N;
    }
    F[V] = (1 << 27);
    sort(F, F + V);
    int ans = 0;
    for (int i = 1; i <= V; i++) {
      if (F[i] - F[i - 1] >= d) {
        ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
