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;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long po2(long long n) {
  long long ans = 0;
  while (n % 128 == 0) {
    ans += 7;
    n /= 128;
  }
  while (n % 2 == 0) {
    ans += 1;
    n /= 2;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n = 20, m = 31, temp;
  vector<unsigned long long> v1, v2;
  set<unsigned long long> v3;
  for (int i = 0; i < m + 1; i++) {
    temp = pow((long long)2, (long long)i);
    v1.push_back(temp);
    v3.insert(temp);
  }
  for (int i = 0; i < n + 1; i++) {
    temp = pow((long long)3, (long long)i);
    v2.push_back(temp);
    v3.insert(temp);
  }
  for (auto it1 : v1) {
    for (auto it2 : v2) {
      v3.insert(it1 * it2);
    }
  }
  long long l, r, count1 = 0;
  cin >> l >> r;
  for (auto it3 : v3) {
    if (it3 >= l && it3 <= r) count1++;
  }
  cout << count1 << "\n";
  return 0;
}
