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;
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long N = 200005, INF = 2000000000000000000;
const long long SIX = 1000000;
bool prime(long long n) {
  long long i = 0;
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long i = 0, j = 0, k = 0;
  long long ans = 0;
  for (i = 0; i < 40; i++) {
    for (j = 0; j < 40; j++) {
      long long kk = 1;
      for (k = 0; k < i; k++) {
        kk = kk * 2;
      }
      for (k = 0; k < j; k++) {
        kk = kk * 3;
        if (kk > 2000000000) break;
      }
      if (l <= kk && kk <= r) ans++;
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}
