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;
const int N = 100001;
int p[N];
int r[N];
int f[N];
const int M = 10001;
int find(int x) {
  if (p[x] == x) {
    return x;
  } else {
    return p[x] = find(p[x]);
  }
}
void unite(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  if (r[x] < r[y]) {
    p[x] = y;
  } else {
    p[y] = x;
    if (r[x] == r[y]) r[x]++;
  }
}
void uft() {
  for (int i = 1; i <= v; i++) {
    p[i] = i;
    r[i] = 0;
  }
  for (int i = 1; i <= v; i++) {
    for (int j = i; j <= v; j++) {
      if (abs(f[i] - f[j]) < d) unite(i, j);
    }
  }
  int ans = 0;
  vector<int> memo(v + 1, 0);
  for (int i = 1; i <= v; i++) {
    int root = find(i);
    if (memo[root] == 0) {
      memo[root] = 1;
      ans++;
    }
  }
  cout << ans << endl;
}
void dfs(int n, int g, vector<int>& groups) {
  queue<int> que;
  que.push(n);
  while (!que.empty()) {
    int p = que.front();
    que.pop();
    if (groups[p] != 0) continue;
    groups[p] = g;
    for (int i = n + 1; i <= v; i++) {
      if (groups[i] == 0 && abs(f[p] - f[i]) < d) {
        que.push(i);
      }
    }
  }
}
void solve() {
  int g = 1;
  vector<int> groups(v + 1, 0);
  for (int i = 1; i <= v; i++) {
    if (groups[i] == 0) {
      dfs(i, g, groups);
      g++;
    }
  }
  cout << (g - 1) << endl;
}
int main() {
  fill(f, f + N, 0);
  f[0] = 1;
  f[1] = 2;
  for (int i = 2; i <= N - 1; i++) {
    f[i] = (f[i - 1] + f[i - 2]) % N;
  }
  long t, elapsed;
  while (cin >> v >> d) {
    uft();
  }
}
