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 <cstdio>
#include <iostream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;

class UnionFindTree
{
    int n;
    vector<int> parent; // em[h
    vector<int> rank;   // ØÌ³ÌãÀ
    vector<int> num;    // O[vÌvf
    int find(int x){
        if(parent[x] == x)
            return x;
        return parent[x] = find(parent[x]);
    }
public:
    UnionFindTree(int n0){ // RXgN^
        n = n0;
        parent.resize(n);
        for(int i=0; i<n; ++i)
            parent[i] = i;
        rank.assign(n, 0);
        num.assign(n, 1);
    }
    void unite(int x, int y){ // xÆyÌO[vð¹
        if((x = find(x)) != (y = find(y))){
            if(rank[x] < rank[y]){
                parent[x] = y;
                num[y] += num[x];
            }else{
                parent[y] = x;
                if(rank[x] == rank[y])
                    ++ rank[x];
                num[x] += num[y];
            }
            -- n;
        }
    }
    bool same(int x, int y){ // xÆyÌO[vª¯¶©ð²×é
        return find(x) == find(y);
    }
    int getNum(){ // O[vÌðÔ·
        return n;
    }
    int getNum(int x){ // xÌO[vÌvfðÔ·
        return num[find(x)];
    }
};

const int n = 1001;

int main()
{
    vector<int> a(n+1);
    a[0] = 1;
    a[1] = 2;
    for(int i=2; i<=n; ++i)
        a[i] = (a[i-1] + a[i-2]) % n;

    for(;;){
        int v, d;
        if(!(cin >> v >> d))
            return 0;

        UnionFindTree uft(v);
        for(int i=1; i<=v; ++i){
            for(int j=i+1; j<=v; ++j){
                if(abs(a[i] - a[j]) < d)
                    uft.unite(i-1, j-1);
            }
        }

        cout << uft.getNum() << endl;
    }
}