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 main() {
  map<int, int> table;
  table[0] = 1;
  table[1] = 2;
  for (int x = 2; x <= 1010; ++x) {
    table[x] = (table[x - 1] % 1001 + table[x - 2] % 1001) % 1001;
  }
  int v, d;
  while (scanf("%d %d", &v, &d), v || d) {
    int tot = 0;
    int graph[v + 1];
    for (int x = 1; x < v + 1; ++x) {
      graph[x] = 0;
    }
    for (int x = 1; x <= v; ++x) {
      for (int y = x + 1; y <= v; ++y) {
        if (abs(table[x] - table[y]) < d) {
          if (graph[x] == 0 && graph[y] == 0) {
            graph[y] = x;
          } else if (graph[x] != 0) {
            graph[y] = x;
          } else {
            graph[x] = y;
          }
        }
      }
    }
    for (int x = 1; x < v + 1; ++x) {
      if (graph[x] == 0) tot++;
      cout << x << ":" << table[x] << ":" << graph[x] << endl;
    }
    cout << tot << endl;
  }
  return 0;
}
