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>
using namespace std;
int v, d, f1, f2;
unsigned long long int f[1010];
int node[1001 + 10][1001 + 10];
void rmNode(int i) {
  node[i][i] = 0;
  for (int j = 0; j < 1001; j++) {
    if (node[j][j] == 0) continue;
    if (node[i][j] == 1) {
      rmNode(j);
    }
  }
}
int main() {
  f[0] = 1;
  f[1] = 2;
  for (int i = 2; i < 1010; i++) {
    f[i] = f[i - 1] + f[i - 2];
  }
  while (cin >> v >> d) {
    memset(node, 0, sizeof(node));
    for (int i = 1; i <= v; i++) {
      f1 = f[i] % 1001;
      node[f1][f1] = 1;
      for (int j = 1; j <= v; j++) {
        f2 = f[j] % 1001;
        if (abs(f1 - f2) < d) {
          node[f1][f2] = 1;
          node[f2][f1] = 1;
        }
      }
    }
    int n = 0;
    for (int i = 0; i < 1001; i++) {
      if (node[i][i] == 0) continue;
      n++;
      rmNode(i);
    }
    cout << n << endl;
  }
  return 0;
}
