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>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define repp(i, n) rep(i, 0, n)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
using namespace std;

const int mod = 1001;
 
signed main(){
    int d[1010] = {0};
    d[0] = 1; d[1] = 2;
    rep(i, 2, 1001){
        d[i] = (d[i - 1] + d[i - 2]) % mod;
    }
    int n, a;
    while(cin >> n >> a){
        int ans = 1;
        set<int> st;
        vector<int> mp;
        rep(i, 1, n + 1){
            if(!st.count(d[i])){
                st.insert(d[i]);
                mp. push_back(d[i]);
            }
        }
        sort(all(mp));
        rep(i, 1, mp.size()){
            if(abs(mp[i] - mp[i - 1]) >= a) ans ++;
        }
        cout << ans << endl;
    }
}