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 <bits/stdc++.h>
int visited[10000], n, M[1001 + 2][1001 + 2];
void dfs(int u) {
  int V;
  visited[u] = 1;
  for (V = 1; V <= n; V++) {
    if (M[u][V] == 1 && visited[V] == 0) dfs(V);
  }
}
int main(void) {
  int V, X, d, i, j, f[10000], count;
  for (i = 0; i < 1001 + 2; i++) {
    for (j = 0; j < 1001 + 2; j++) {
      M[i][j] = 0;
    }
  }
  while (scanf("%d %d", &V, &d) != EOF) {
    n = V;
    f[0] = 1;
    f[1] = 2;
    count = 0;
    for (i = 2; i <= V + 1; i++) {
      f[i] = (f[i - 1] + f[i - 2]) % 1001;
    }
    for (i = 1; i <= V; i++) {
      for (j = 1; j <= V; j++) {
        X = f[i] - f[j];
        if (abs(X) < d && i != j) {
          M[f[i]][f[j]] = 1;
          M[f[j]][f[i]] = 1;
        } else {
          M[f[i]][f[j]] = 0;
          M[f[j]][f[i]] = 0;
        }
      }
    }
    for (i = 0; i < 1002; i++) {
      visited[i] = 0;
    }
    for (i = 1; i <= V; i++) {
      if (visited[i] != 1) {
        count++;
        dfs(i);
      }
    }
    printf("%d\n", count);
  }
  return 0;
}
