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?

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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 <algorithm>
#include <vector>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>

using namespace std;

typedef long long ll;

const int N = 1001;
int V, d;

ll f[N + 1];

int main(void) {
    f[0] = 1L; f[1] = 2L;
    for(int i = 2; i <= N; i++) {
        f[i] = (ll) ((f[i - 2] % N) + (f[i - 1] % N)) % N;
    }
    while(~scanf("%d %d", &V, &d)) {
        vector<ll> tmp;
        for(int i = 1; i <= V; i++) {
            tmp.push_back((ll)f[i]);
        }
        int ans = 1; sort(tmp.begin(), tmp.end());
        for(int i = 1; i < (int) tmp.size(); i++) {
            if(tmp[i] - tmp[i - 1] >= d) ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}