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;
long long a[10000];
long long pow(int x, int i) {
  long long ans = 1;
  while (i) {
    if (i & 1) ans *= x;
    i /= 2;
    x = x * x;
  }
  return ans;
}
int main() {
  int count = 0;
  for (int i = 0; i <= 31; i++)
    for (int j = 0; j <= 20; j++) {
      long long ans = 1;
      ans *= pow(2, i);
      ans *= pow(3, j);
      if (ans <= 2000000000) a[count++] = ans;
    }
  sort(a, a + count);
  int l, r;
  while (cin >> l >> r) {
    int l1 = 0, r1 = count - 1, ok = 0;
    if (l > a[count - 1]) ok = 1;
    for (int i = 0; i < count; i++) {
      if (l > a[i] && r < a[i + 1]) ok = 1;
    }
    if (ok) {
      cout << 0 << endl;
      continue;
    }
    for (int i = 0; i < count; i++)
      if (a[i] >= l) {
        l1 = i;
        break;
      }
    for (int i = 0; i < count; i++)
      if (a[i] > r) {
        r1 = i - 1;
        break;
      }
    cout << (r1 - l1 + 1) << endl;
  }
}
