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 int maxn = 1e5 + 10;
const long long int MOD = 1e9 + 7;
vector<long long int> numbers;
long long int POW(int a, int b) {
  long long int ans = 1;
  for (int i = 0; i < b; i++) ans *= a;
  return ans;
}
void pre_cal() {
  long long int two[34];
  long long int three[34];
  two[0] = three[0] = 1;
  for (int i = 1; i < 32; i++) {
    two[i] = two[i - 1] * 2;
    three[i] = three[i - 1] * 3;
  }
  for (int i = 0; i < 32; i++) {
    for (int j = 0; j < (32 - i); j++) {
      long long int t = two[i] * three[j];
      if (t < 10000000000) numbers.push_back(t);
    }
  }
  sort(numbers.begin(), numbers.end());
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  pre_cal();
  long long int l, r;
  cin >> l >> r;
  int x = upper_bound(numbers.begin(), numbers.end(), l) - numbers.begin();
  int y = lower_bound(numbers.begin(), numbers.end(), r) - numbers.begin();
  if (numbers[y] == r) y++;
  if (numbers[x - 1] == l) x--;
  cout << (y - x) << '\n';
  return 0;
}
