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;
using ll = long long;
ll max(ll a, ll b) { return (a > b ? a : b); }
ll min(ll a, ll b) { return (a > b ? b : a); }
ll a, b, c, d, e, n, m, k, q, x, y, z, w, t, cur, cnt, ma,
    mi = LONG_LONG_MAX, l, r, mid, sum, i, j, pro = 1, f;
ll arr[200005];
ll arr1[200005];
ll arr2[200005];
ll dp[200005];
ll arr3[1000][1000];
vector<ll> v[2000];
string s;
int dfs(int i) {
  arr[i] = 1;
  int mi = dp[i];
  for (auto c : v[i]) {
    if (!arr[c]) {
      mi = min(mi, dfs(c));
    }
  }
  return mi;
}
void solve() {
  cin >> l >> r;
  vector<ll> v;
  while (pro < INT_MAX) {
    v.push_back(pro);
    pro = min(v[a] * 2, v[b] * 3);
    if (pro == v[a] * 2) a++;
    if (pro == v[b] * 3) b++;
  }
  for (auto c : v)
    if (c <= r && c >= l) cnt++;
  cout << cnt;
}
int main() {
  ll t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
