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;
void solve() {
  long long l, r;
  cin >> l >> r;
  long long two = 1, three = 3, cnt = 0, no_tw = 0, no_th = 0;
  while (two <= r) {
    if (two >= l && two <= r) {
      cnt++;
      no_tw++;
      two *= 2;
    } else if (two > r) {
      break;
    } else {
      two *= 2;
      no_tw++;
    }
  }
  while (three <= r) {
    if (three >= l && three <= r) {
      cnt++;
      no_th++;
      three *= 3;
    } else if (three > r) {
      break;
    } else {
      three *= 3;
      no_th++;
    }
  }
  no_th++;
  long long tw = 2, th = 3, fx2 = 1, fx3 = 3;
  for (int i = 0; i < no_tw + 10; i++) {
    fx2 *= 2;
    th = 3;
    for (int j = 0; j < no_th + 10; j++) {
      long long p = fx2 * th;
      if (p >= l && p <= r) {
        cnt++;
      }
      th *= 3;
    }
  }
  cout << cnt;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}
