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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int MOD = 1001;
const int MAXV = 1000;

struct UnionFind {
    vector<int> data;
    int __size;
    UnionFind(int size) : data(size, -1), __size(size) { }
    void unionSet(int x, int y) {
        if ((x = root(x)) != (y = root(y))) {
            if (data[y] < data[x]) swap(x, y);
            data[x] += data[y]; data[y] = x; __size--;
        }
    }
    bool findSet(int x, int y) { return root(x) == root(y); }
    int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
    int size(int x) { return -data[root(x)]; }
    int size() { return __size; }
    vector<int> subset_roots() {
        vector<int> roots;
        for (int i = 0; i < data.size(); i++) if (data[i] < 0) roots.push_back(i);
        return roots;
    }
};

int main() {
    vector<int> F(MAXV + 1);
    F[0] = 1, F[1] = 2;
    for (int i = 2; i <= MAXV; i++) (F[i] = F[i - 1] + F[i - 2]) %= MOD;

    int V, d;
    while (cin >> V >> d) {
        UnionFind uf = UnionFind(V + 1);    // 1-origin
        for (int i = 1; i <= V; i++) {
            for (int j = i + 1; j <= V; j++) {
                if (abs(F[i] - F[j]) < d) {
                    uf.unionSet(i, j);
                }
            }
        }
        cout << uf.size() - 1 << endl;   // 1-origin
    }
    return 0;
}
