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;
const int N = 1e5 + 4;
map<string, string> maap;
vector<int> circle;
bool vis[N];
int parent[N];
int chk[N];
int color[N];
int tim[2][N];
int dis[N];
int position[N];
vector<int> adj[N];
vector<int> adj1[N];
vector<int> graph[N];
bool has_cycle;
int maxdis, maxnode, Totnode, depth = 1;
bool ok;
queue<int> q;
stack<int> stk;
vector<int> solution;
int indegree[N];
int go[N];
int to[N];
map<int, int> mp1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  int cas = 0;
  ll n, m, i, j, cnt1 = 0, cnt2 = 0, cnt = 0, even = 0, odd = 0, len, k, r, l,
                 z = 0, x = 0, y = 0, flag = 0, sum = 0;
  ll a = 0, b = 0, c = 0, d = 0, ans = 0LL, rem, quot, zero = 0, fst = 0,
     null = 0, snd = 0, lst = 0, rone = 0, one = 0, pos = 0, neg = 0,
     mn = INT_MAX, mx = INT_MIN;
  char ch;
  int h1, h2, m1, m2, h;
  int velo1, velo2, ac1, ac2, time;
  int node, edge, u, v, cost;
  string str, str1 = "", str2 = "", str3 = "", strstore1 = "", strstore2 = "";
  int best, worst;
  cin >> l >> r;
  ll arr2[50];
  arr2[0] = 1;
  if (arr2[0] >= l) {
    cnt++;
  }
  for (int i = 1; i <= 31; i++) {
    arr2[i] = arr2[i - 1] * 2;
    if (arr2[i] >= l && arr2[i] <= r) {
      cnt++;
    }
  }
  ll arr3[30];
  arr3[0] = 1;
  for (int i = 1; i <= 25; i++) {
    arr3[i] = arr3[i - 1] * 3;
    if (arr3[i] >= l && arr3[i] <= r) {
      cnt++;
    }
  }
  for (int i = 1; i <= 31; i++) {
    for (int j = 1; j <= 25; j++) {
      x = arr2[i] * arr3[j];
      if (x >= l && x <= r) {
        cnt++;
      }
    }
  }
  cout << cnt << endl;
  return 0;
}
