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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
const int mod = 1e9 + 7;
using namespace std;
template <class T>
void cinn(T& x) {
  cin >> x;
}
template <class H, class... T>
void cinn(H& h, T&... t) {
  cinn(h);
  cinn(t...);
}
template <class A>
void cinn(vector<A>& x) {
  for (auto& a : x) cinn(a);
}
long long bpow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 0) {
      a = (a * a * 1LL);
      b = (b) / 2;
    } else {
      ans = (ans * a * 1LL);
      b--;
    }
  }
  return ans;
}
void Shazam() {
  long long l, r;
  cinn(l, r);
  long long cnt = 0;
  for (long long i = 0; i < 40; i++) {
    for (long long j = 0; j < 40; j++) {
      long long mid = (bpow(2, i) * 1LL * bpow(3, j));
      if (mid >= l && mid <= r) cnt++;
      if (mid > r) break;
    }
  }
  cout << cnt << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    Shazam();
  }
  return 0;
}
