A positive integer is called a 2-3-integer, if it is equal to 2x·3y for some non-negative integers x and y. In other words, these integers are such integers that only have 2 and 3 among their prime divisors. For example, integers 1, 6, 9, 16 and 108 — are 2-3 integers, while 5, 10, 21 and 120 are not.

Print the number of 2-3-integers on the given segment [l, r], i. e. the number of sich 2-3-integers t that l ≤ t ≤ r.

Input

The only line contains two integers l and r (1 ≤ l ≤ r ≤ 2·109).

Output

Print a single integer the number of 2-3-integers on the segment [l, r].

Examples

Input

1 10


Output

7


Input

100 200


Output

5


Input

1 2000000000


Output

326

Note

In the first example the 2-3-integers are 1, 2, 3, 4, 6, 8 and 9.

In the second example the 2-3-integers are 108, 128, 144, 162 and 192.
import java.util.HashSet;
import java.util.Scanner;
        
public class Codeforces {    
    public static void main(String[] args) {
        TaskA Solver = new TaskA();        
        Solver.Solve();        
    }       
    private static class TaskA {                          
        private void Solve() {    
            Scanner in = new Scanner(System.in);
            long l = in.nextInt();
            long r = in.nextInt();            
            final int MAX = 32;
            long ans = 0;            
            HashSet <Long> set = new HashSet();
            for (int p1 = 0; p1 < MAX; p1++) {
                for (int p2 = 0; p2 < MAX; p2++) {
                    long m1 = BinPow(2, p1);
                    long m2 = BinPow(3, p2);
                    long m = m1 * m2;
                    if (m >= l && m <= r) {
                        if (!set.contains(m)) {
                            ans++;
                            set.add(m);                            
                        }                        
                    }
                }                
            }
            System.out.println(ans);
        }    
        private long BinPow(long a, long n) {
            long res = 1;
            while (n > 0) {
                if ((n & 1) == 1)
                    res *= a;
                a *= a;
                n >>= 1;
            }
            return res;
        }        
    }
}