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;
const double PI = acos(-1);
const double EPS = 1e-9;
void task();
int main() {
  ios_base::sync_with_stdio(false);
  srand(time(0));
  task();
  return 0;
}
const long long MAX = 2e9;
int l = 0;
int r = 0;
void init() { cin >> l >> r; }
vector<long long> v;
set<long long> s;
void dfs(long long a) {
  v.push_back(a);
  s.insert(a);
  if (a * 2 <= MAX && s.find(a * 2) == s.end()) dfs(a * 2);
  if (a * 3 <= MAX && s.find(a * 3) == s.end()) dfs(a * 3);
}
void solve() {
  dfs(1);
  sort(v.begin(), v.end());
}
void print_answer() {
  auto right = lower_bound(v.begin(), v.end(), r);
  auto left = lower_bound(v.begin(), v.end(), l);
  int ans = right - left;
  if (*right == r) ++ans;
  cout << ans;
}
void task() {
  init();
  solve();
  print_answer();
}
