At first, let's define function f(x) as follows: $$$ \begin{matrix} f(x) & = & \left\{ \begin{matrix} x/2 & \mbox{if } x  is even \\\ x - 1 & \mbox{otherwise } \end{matrix} \right. \end{matrix} $$$

We can see that if we choose some value v and will apply function f to it, then apply f to f(v), and so on, we'll eventually get 1. Let's write down all values we get in this process in a list and denote this list as path(v). For example, path(1) = [1], path(15) = [15, 14, 7, 6, 3, 2, 1], path(32) = [32, 16, 8, 4, 2, 1].

Let's write all lists path(x) for every x from 1 to n. The question is next: what is the maximum value y such that y is contained in at least k different lists path(x)?

Formally speaking, you need to find maximum y such that \left| \{ x ~|~ 1 ≤ x ≤ n, y ∈ path(x) \} \right| ≥ k.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 10^{18}).

Output

Print the only integer — the maximum value that is contained in at least k paths.

Examples

Input


11 3


Output


5


Input


11 6


Output


4


Input


20 20


Output


1


Input


14 5


Output


6


Input


1000000 100


Output


31248

Note

In the first example, the answer is 5, since 5 occurs in path(5), path(10) and path(11).

In the second example, the answer is 4, since 4 occurs in path(4), path(5), path(8), path(9), path(10) and path(11).

In the third example n = k, so the answer is 1, since 1 is the only number occuring in all paths for integers from 1 to 20.
#include <bits/stdc++.h>
using namespace std;
int MOD = 1e9 + 7;
unsigned long long int power(unsigned long long int a,
                             unsigned long long int b) {
  unsigned long long int res = 1;
  a = a % MOD;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % MOD;
      b--;
    }
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
unsigned long long int fermat_inv(unsigned long long int y) {
  return power(y, MOD - 2);
}
unsigned long long int gcd(unsigned long long int a, unsigned long long int b) {
  return (b == 0) ? a : gcd(b, a % b);
}
unsigned long long int min(unsigned long long int a, unsigned long long int b) {
  return (a > b) ? b : a;
}
unsigned long long int max(unsigned long long int a, unsigned long long int b) {
  return (a > b) ? a : b;
}
bool prime[1000001];
vector<int> primes;
void SieveOfEratosthenes(int n) {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p < 1000001; p++)
    if (prime[p]) primes.push_back(p);
}
unsigned long long int fact[100010];
unsigned long long int finv[100010];
void factorial(int n) {
  fact[0] = 1;
  finv[0] = 1;
  for (int i = 1; i <= n; i++)
    fact[i] = fact[i - 1] * i, fact[i] %= MOD, finv[i] = fermat_inv(fact[i]);
}
unsigned long long int ncr(unsigned long long int n, unsigned long long int r) {
  if (n < r)
    return 0;
  else {
    unsigned long long int x = finv[r] * finv[n - r] % MOD;
    return fact[n] * x % MOD;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int te = 1;
  while (te--) {
    unsigned long long int n, k;
    cin >> n >> k;
    unsigned long long int l = 1, r = (n + 1) / 2, ans = 0;
    while (l <= r) {
      unsigned long long int mid = (l + r) / 2, x = 2 * mid - 1;
      unsigned long long int cnt = 1, j = 2;
      while (j * x <= n) {
        if (n > j * x + j - 1)
          cnt += j;
        else
          cnt += n - j * x + 1;
        j *= 2;
      }
      if (cnt >= k)
        l = mid + 1, ans = max(ans, x);
      else
        r = mid - 1;
    }
    l = 1, r = n / 2;
    while (l <= r) {
      unsigned long long int mid = (l + r) / 2, x = 2 * mid;
      unsigned long long int cnt = 1 + (x + 1 <= n), j = 2;
      while (j * x <= n) {
        if (n > j * x + 2 * j - 1)
          cnt += 2 * j;
        else
          cnt += n - j * x + 1;
        j *= 2;
      }
      if (cnt >= k)
        l = mid + 1, ans = max(ans, x);
      else
        r = mid - 1;
    }
    cout << ans;
  }
}
