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 <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
#include <utility>

using namespace std;

int maxn = 1000;
int n = 1001;
int tree[1000];
int fib[1000];

int root(int a);
bool is_same_set(int p, int q);
void joint(int x, int y);
void makefib(int maxn);

int main()
{
    makefib(1000);
    int v, d, res;
    while (cin >> v)
    {
        if (cin.eof())
            break;
        cin >> d;

        for (int i = 0; i < maxn; i++)
        {
            tree[i] = i;
        }
        for (int i = 0; i < v; i++)
        {
            for (int j = i; j < v; j++)
            {
                if (abs(fib[i] - fib[j]) < d)
                {
                    joint(i, j);
                }
            }
        }
        res = 0;
        for (int i = 0; i < v; i++)
        {
            if (root(i) == i)
            {
                res++;
            }
        }
        cout << res << endl;
    }

    return 0;
}

void makefib(int maxn)
{
    fib[0] = 2;
    fib[1] = 3;

    for (int i = 2; i < maxn; i++)
    {
        fib[i] = (fib[i - 1] + fib[i - 2]) % n;
    }
}

int root(int a)
{
    if (tree[a] == a)
    {
        return a;
    }
    else
    {
        tree[a] = root(tree[a]);
        return tree[a];
    }
}

bool is_same_set(int p, int q)
{
    return root(p) == root(q);
}

void joint(int x, int y)
{
    tree[root(x)] = root(y);
}

