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.
#include <bits/stdc++.h>
using namespace std;
long long int _xor(long long int a, long long int b) { return a ^ b; }
long long int _and(long long int a, long long int b) { return a & b; }
long long int _or(long long int a, long long int b) { return a | b; }
long long int _not(long long int a) { return ~a; }
long long int cum[1000000], pi[2005];
int main() {
  long long int b, c = 0, x, y, z, t, i, j, k, l1, l2, m, n, c1 = 0, c2 = 0, q,
                   poisa, f = 0, ans = 0, a;
  scanf("%lld", &l1);
  scanf("%lld", &l2);
  for (i = 1; i <= l1 - 1; i = i * 2) {
    for (j = 1; j <= l1 - 1; j = j * 3) {
      if (i * j > l1 - 1) {
        f = 1;
        break;
      } else if (1 <= i * j && i * j <= l1 - 1) {
        c1++;
      }
    }
  }
  for (i = 1; i <= l2; i = i * 2) {
    for (j = 1; j <= l2; j = j * 3) {
      if (i * j > l2) {
        f = 1;
        break;
      } else if (1 <= i * j && i * j <= l2) {
        c2++;
      }
    }
  }
  cout << c2 - c1;
  return 0;
}
