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 <sstream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <iostream>
#include <utility>
#include <set>
#include <cctype>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
using namespace std;
#define INF 100000000

typedef long long ll;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};

const int MAXN = 1010;
const int MOD = 1001;

int F[MAXN];

void init() {
    F[0] = 1;
    F[1] = 2;
    for (int i = 2; i < MAXN; i++) {
        F[i] = (F[i-1] + F[i-2]) % MOD;
        if (F[i] < 0) F[i] += MOD;
    }
}

int par[MAXN];
void union_init() {
    for (int i = 0; i < MAXN; i++) par[i] = i;
}

int find(int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
}

void unite(int x, int y) {
    x = find(x); y = find(y);
    if (x == y) return;
    par[x] = y;
}

bool same(int x, int y) {
    return find(x) == find(y);
}

int main(void) {
    init();
    int V, d;
    while (cin >> V >> d) {
        union_init();
        for (int i = 1; i <= V; i++) {
            for (int j = i+1; j <= V; j++) {
                if (abs(F[i] - F[j]) < d) {
                    unite(i, j);
                }
            }
        }
        int ans = 0;
        for (int i = 1; i <= V; i++) {
            if (par[i] == i) ans++;
        }
        cout << ans << endl;
    }
    return 0;
}