#include <bits/stdc++.h>
using namespace std;
map<long long int, long long int> v;
int main() {
  v.insert(make_pair(2, 1));
  long long int a, b, c, i, j, count = 0, val, vali, temp, counter = 2, ans = 0;
  cin >> a >> b;
  vali = 2;
  count = 2;
  while (vali <= log2(1e18) + 1) {
    val = pow(2, vali) - 1;
    temp = 1;
    vali++;
    long long int up = pow(2, vali - 2);
    while (temp < up) {
      v.insert(make_pair(val - temp, counter));
      temp *= 2;
      counter++;
    }
  }
  map<long long int, long long int>::iterator itr;
  for (itr = v.begin(); itr != v.end(); itr++) {
    if (itr->first >= a && itr->first <= b) ans++;
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

The year 2015 is almost over.

Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510 = 111110111112. Note that he doesn't care about the number of zeros in the decimal representation.

Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?

Assume that all positive integers are always written without leading zeros.

Input

The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 1018) — the first year and the last year in Limak's interval respectively.

Output

Print one integer – the number of years Limak will count in his chosen interval.

Examples

Input

5 10


Output

2


Input

2015 2015


Output

1


Input

100 105


Output

0


Input

72057594000000000 72057595000000000


Output

26

Note

In the first sample Limak's interval contains numbers 510 = 1012, 610 = 1102, 710 = 1112, 810 = 10002, 910 = 10012 and 1010 = 10102. Two of them (1012 and 1102) have the described property.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
map<long long int, long long int> v;
int main() {
  v.insert(make_pair(2, 1));
  long long int a, b, c, i, j, count = 0, val, vali, temp, counter = 2, ans = 0;
  cin >> a >> b;
  vali = 2;
  count = 2;
  while (vali <= log2(1e18) + 1) {
    val = pow(2, vali) - 1;
    temp = 1;
    vali++;
    long long int up = pow(2, vali - 2);
    while (temp < up) {
      v.insert(make_pair(val - temp, counter));
      temp *= 2;
      counter++;
    }
  }
  map<long long int, long long int>::iterator itr;
  for (itr = v.begin(); itr != v.end(); itr++) {
    if (itr->first >= a && itr->first <= b) ans++;
  }
  cout << ans;
  return 0;
}

```