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 long long int INF = 9223372036854775807LL;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroint-loops")
long long int power(long long int a, long long int b, long long int m) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = ((res % m) * (a % m)) % m;
    a = ((a % m) * (a % m)) % m;
    b >>= 1;
  }
  return res;
}
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long int l, r;
  cin >> l >> r;
  vector<long long int> v;
  for (long long int i = 0; i <= 20; i++) {
    for (long long int j = 0; j <= 31; j++) {
      long long int x = power(3, i);
      long long int y = power(2, j);
      v.push_back(x * y);
    }
  }
  long long int ans = 0;
  for (auto i : v) {
    if (i >= l and i <= r) ans++;
  }
  cout << ans;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tt = 1;
  while (tt--) {
    solve();
    cout << "\n";
  }
  return 0;
}
