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>
using V = vector<T>;
template <typename T>
using VV = vector<V<T>>;
template <typename T>
istream& operator>>(istream& cin, V<T>& arr) {
  for (T& val : arr) {
    cin >> val;
  }
  return cin;
}
template <typename T>
ostream& operator<<(ostream& cout, V<T>& arr) {
  for (T& val : arr) {
    cout << val << " ";
  }
  return cout;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int maxPow2 = 32;
  int maxPow3 = 21;
  vector<vector<long long>> nums(maxPow3, vector<long long>(maxPow2));
  nums[0][0] = 1;
  for (long long i = 1; i < maxPow2; i++) nums[0][i] = nums[0][i - 1] * 2;
  for (long long i = 1; i < maxPow3; i++) nums[i][0] = nums[i - 1][0] * 3;
  for (long long i = 1; i < maxPow3; i++) {
    for (long long j = 1; j < maxPow2; j++) {
      nums[i][j] = nums[i][0] * nums[0][j];
    }
  }
  long long l, r;
  cin >> l >> r;
  long long cnt_l = 0;
  long long cnt_r = 0;
  for (long long i = 0; i < maxPow3; i++) {
    for (long long j = 0; j < maxPow2; j++) {
      cnt_l += nums[i][j] < l;
      cnt_r += nums[i][j] <= r;
    }
  }
  cout << cnt_r - cnt_l << '\n';
  return 0;
}
