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;
bool p[10000000];
void seive() {
  int i, j;
  for (i = 2; i * i <= 10000000; i++) {
    if (!p[i]) {
      for (j = i * i; j <= 10000000; j += i) p[j] = true;
    }
  }
}
long long int big_mod(long long int b, long long int p, long long int m) {
  long long int ans = 1;
  while (p) {
    if (p % 2 == 1) {
      ans = (b % m * ans % m) % m;
    }
    p = p >> 1;
    b = (b * b) % m;
  }
  return ans;
}
int main() {
  long long int mod, t, k, i, m, n, x, j = 0, l, c = 0, r = 0, w, s, d = 0,
                                       f = 0, sum = 0, mul, mx1, mx, s1 = 0,
                                       s2 = 0;
  cin >> l >> r;
  for (i = 0; i <= 31; i++) {
    d = big_mod(2, i, 1000000000000);
    for (j = 0; j <= 20; j++) {
      c = big_mod(3, j, 1000000000000);
      if (c * d >= l && c * d <= r) {
        sum++;
      }
    }
  }
  cout << sum << endl;
  return 0;
}
