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>
const int inf = INT_MAX;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 111;
using namespace std;
long long l, r, a[55], b[55];
vector<long long> v;
int main() {
  long long k = 2;
  cin >> l >> r;
  v.push_back(1ll);
  for (int i = 1; i <= 30; i++) {
    a[i] = k;
    if (k > 3 * 1e9) break;
    v.push_back(k);
    k *= 2;
  }
  k = 3;
  for (int i = 1; i <= 30; i++) {
    b[i] = k;
    if (k > 3 * 1e9) break;
    v.push_back(k);
    k *= 3;
  }
  for (int i = 1; i <= 29; i++)
    for (int j = 1; j <= 29; j++) {
      if (a[i] * b[j] > 3 * 1e9) break;
      v.push_back(a[i] * b[j]);
    }
  sort(v.begin(), v.end());
  int x = 0, y = (int)v.size() - 1, pos1, pos2;
  while (x <= y) {
    long long mid = (x + y) >> 1;
    if (v[mid] >= l) {
      pos1 = mid;
      y = mid - 1;
    } else
      x = mid + 1;
  }
  x = 0, y = (int)v.size() - 1;
  while (x <= y) {
    long long mid = (x + y) >> 1;
    if (v[mid] <= r) {
      pos2 = mid;
      x = mid + 1;
    } else
      y = mid - 1;
  }
  cout << pos2 - pos1 + 1;
}
