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;
set<int> todos;
void calc(int lim, int a, int b, bool prend) {
  if (2000000000 < 1ll * pow(2, a) * pow(3, b)) {
    return;
  }
  todos.insert(pow(2, a) * pow(3, b));
  if (prend) {
    calc(lim, a, b + 1, true);
  } else {
    calc(lim, a + 1, b, false);
    calc(lim, a, b + 1, true);
  }
}
int main() {
  int a, b;
  cin >> a >> b;
  vector<long long> v2, v3;
  long long x = 1;
  for (int i = 0; i <= 30; i++) {
    v2.push_back(x);
    x *= 2;
  }
  x = 1;
  for (int i = 0; i <= 19; i++) {
    v3.push_back(x);
    x *= 3;
  }
  for (int i = 0; i < v2.size(); i++) {
    for (int j = 0; j < v3.size(); j++) {
      if (v2[i] * v3[j] <= 2e9) {
        todos.insert(v2[i] * v3[j]);
      }
    }
  }
  int c = 0;
  auto it = lower_bound(todos.begin(), todos.end(), a);
  for (; *it <= b && it != todos.end(); it++) {
    c++;
  }
  cout << c;
}
