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;
struct Matrix2x2 {
  int a, b, c, d;
};
Matrix2x2 mult(Matrix2x2 A, Matrix2x2 B) {
  Matrix2x2 C = {0};
  C.a = ((A.a * B.a) % 1001 + (A.b * B.c) % 1001) % 1001;
  C.b = ((A.a * B.b) % 1001 + (A.b * B.d) % 1001) % 1001;
  C.c = ((A.c * B.a) % 1001 + (A.d * B.c) % 1001) % 1001;
  C.d = ((A.c * B.b) % 1001 + (A.d * B.d) % 1001) % 1001;
  return C;
}
Matrix2x2 expt(Matrix2x2 A, int p) {
  if (p == 1) {
    return A;
  } else if (p % 2 == 1) {
    Matrix2x2 T = expt(A, p - 1);
    return mult(A, T);
  } else {
    Matrix2x2 T = expt(A, p / 2);
    return mult(T, T);
  }
}
int p[1010];
void init(int nn) {
  for (int ii = 0; ii < nn + 1; ii++) {
    p[ii] = ii;
  }
}
int root(int a) {
  if (p[a] == a)
    return a;
  else
    return (p[a] = root(p[a]));
}
void unite(int a, int b) { p[root(a)] = root(b); }
int main() {
  int f[1010];
  for (int k = 0; k < 1010; k++) {
    if (k == 0) {
      f[k] = 1;
    } else if (k == 1) {
      f[k] = 2;
    } else {
      Matrix2x2 A = {1, 1, 1, 0};
      Matrix2x2 G = expt(A, k - 1);
      f[k - 2] = G.a;
    }
  }
  int v, d;
  while (cin >> v >> d) {
    int j = 0;
    init(v);
    for (int i = 1; i <= v; i++) {
      for (int j = i + 1; j <= v; j++) {
        if (abs(f[i] - f[j]) < d) {
          unite(i, j);
        }
      }
    }
    for (int r = 1; r <= v; r++) {
      if (p[r] == r) {
        j = j + 1;
      }
    }
    cout << j << endl;
  }
  return 0;
}
