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 <vector>
#include <algorithm>
using namespace std;

#define SIZE 1001

int main() {
    vector<long long> fibonaccis(SIZE + 1);
    fibonaccis[0] = 1;
    fibonaccis[1] = 2;
    for (int i = 2; i <= SIZE; i++) {
        fibonaccis[i] = (fibonaccis[i - 1] + fibonaccis[i - 2]) % SIZE;
    }
    int v, d;
    while (cin >> v >> d) {
        vector<int> nums(fibonaccis.begin() + 1, fibonaccis.begin() + v + 1);
        sort(nums.begin(), nums.end());
        int result = 1;
        for (int i = 1, size = nums.size(); i < size; i++) {
            if (nums[i] - nums[i - 1] >= d) {
                result++;
            }
        }
        cout << result << endl;
    }
}