#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) {
  long long g = gcd(a, b);
  long long ans = (a * b) / g;
  return ans;
}
vector<long long> primes;
bool prime[10005];
void seive() {
  memset(prime, 1, sizeof(prime));
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 2; i <= 10000; i++) {
    if (prime[i] == 1) {
      for (long long j = i * i; j <= 10000; j += i) prime[j] = 0;
    }
  }
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans % 1000000007 * a % 1000000007) % 1000000007;
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return ans;
}
template <typename T>
std::string NumberToString(T Number) {
  std::ostringstream ss;
  ss << Number;
  return ss.str();
}
long long gcdExtended(long long a, long long b, long long *x, long long *y);
long long modInverse(long long b, long long m) {
  long long x, y;
  long long g = gcdExtended(b, m, &x, &y);
  if (g != 1) return -1;
  return (x % m + m) % m;
}
long long modDivide(long long a, long long b, long long m) {
  a = a % m;
  long long inv = modInverse(b, m);
  return (inv * a) % m;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long ct_bits(long long n) {
  long long ans = 0;
  while (n) {
    ans++;
    n /= 2;
  }
  return ans;
}
void solve() {
  long long a, b;
  cin >> a >> b;
  long long i, totala = ct_bits(a), totalb = ct_bits(b);
  if (totala == totalb) {
    long long num = (1LL << totala) - 1, ans = 0;
    for (i = 0; i < totala - 1; i++) {
      long long curr = num ^ (1LL << i);
      if (curr >= a and curr <= b) {
        ans++;
      }
    }
    cout << ans;
  } else {
    long long ans1 = 0;
    long long ans2 = 0;
    long long ans3 = 0;
    for (long long i = totala + 1; i < totalb; i++) {
      ans3 += i - 1;
    }
    long long num = (1LL << totala) - 1;
    for (i = 0; i < totala - 1; i++) {
      long long curr = num ^ (1LL << i);
      if (curr >= a) {
        ans1++;
      }
    }
    num = (1LL << totalb) - 1;
    for (i = 0; i < totalb - 1; i++) {
      long long curr = num ^ (1LL << i);
      if (curr <= b) {
        ans2++;
      }
    }
    cout << ans1 + ans2 + ans3;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  bool codechef = 0;
  long long t = 1;
  if (codechef) {
    cin >> t;
  }
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) {
  long long g = gcd(a, b);
  long long ans = (a * b) / g;
  return ans;
}
vector<long long> primes;
bool prime[10005];
void seive() {
  memset(prime, 1, sizeof(prime));
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 2; i <= 10000; i++) {
    if (prime[i] == 1) {
      for (long long j = i * i; j <= 10000; j += i) prime[j] = 0;
    }
  }
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans % 1000000007 * a % 1000000007) % 1000000007;
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return ans;
}
template <typename T>
std::string NumberToString(T Number) {
  std::ostringstream ss;
  ss << Number;
  return ss.str();
}
long long gcdExtended(long long a, long long b, long long *x, long long *y);
long long modInverse(long long b, long long m) {
  long long x, y;
  long long g = gcdExtended(b, m, &x, &y);
  if (g != 1) return -1;
  return (x % m + m) % m;
}
long long modDivide(long long a, long long b, long long m) {
  a = a % m;
  long long inv = modInverse(b, m);
  return (inv * a) % m;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long ct_bits(long long n) {
  long long ans = 0;
  while (n) {
    ans++;
    n /= 2;
  }
  return ans;
}
void solve() {
  long long a, b;
  cin >> a >> b;
  long long i, totala = ct_bits(a), totalb = ct_bits(b);
  if (totala == totalb) {
    long long num = (1LL << totala) - 1, ans = 0;
    for (i = 0; i < totala - 1; i++) {
      long long curr = num ^ (1LL << i);
      if (curr >= a and curr <= b) {
        ans++;
      }
    }
    cout << ans;
  } else {
    long long ans1 = 0;
    long long ans2 = 0;
    long long ans3 = 0;
    for (long long i = totala + 1; i < totalb; i++) {
      ans3 += i - 1;
    }
    long long num = (1LL << totala) - 1;
    for (i = 0; i < totala - 1; i++) {
      long long curr = num ^ (1LL << i);
      if (curr >= a) {
        ans1++;
      }
    }
    num = (1LL << totalb) - 1;
    for (i = 0; i < totalb - 1; i++) {
      long long curr = num ^ (1LL << i);
      if (curr <= b) {
        ans2++;
      }
    }
    cout << ans1 + ans2 + ans3;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  bool codechef = 0;
  long long t = 1;
  if (codechef) {
    cin >> t;
  }
  while (t--) {
    solve();
  }
  return 0;
}

```