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;
long long numbit(long long x) { return (1LL << x) - 1LL; }
long long getbit(long long x, long long i) { return (x >> i) & 1LL; }
long long onbit(long long x, long long i) { return x | (1LL << i); }
long long offbit(long long x, long long i) { return x & (~(1LL << i)); }
int newMaskNULL(int mask, int m) { return (mask << 1) % (1 << m); }
int newMask(int mask, int m) { return ((mask << 1) % (1 << m)) | 1; }
const long long Inf64 = 1e18;
const long long Inf = 1e9;
const int Nmax = 1e5 + 10;
const long long Mod9 = 1e9 + 9;
const long long Mod7 = 1e9 + 7;
int main() {
  long long n, k;
  cin >> n >> k;
  long long odd, even;
  even = n / 2;
  odd = n - even;
  long long l, r, mid;
  l = 1;
  r = odd;
  long long r1, r2;
  r1 = -1;
  while (l <= r) {
    mid = (l + r) / 2;
    long long d = 0;
    long long tmp = 1;
    bool ok = true;
    while (ok) {
      if (((1 + (mid - 1) * 1LL * 2) * 1LL * tmp) > n) break;
      if ((1 + (mid - 1) * 1LL * 2) * 1LL * tmp + tmp - 1 <= n)
        d += tmp;
      else {
        d += n - (1 + (mid - 1) * 2) * tmp + 1;
        break;
      }
      tmp *= 2;
      if (d >= k) break;
    }
    if (d >= k) {
      r1 = 1 + (mid - 1) * 2;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  l = 1;
  r = even;
  r2 = -1;
  while (l <= r) {
    mid = (l + r) / 2;
    long long d = 0;
    long long tmp = 1;
    bool ok = true;
    while (ok) {
      if (2 * 1LL * mid * 1LL * tmp > n) break;
      if (2 * 1LL * mid * 1LL * tmp + 2 * 1LL * tmp - 1 <= n)
        d += 2 * tmp;
      else {
        d += n - 2 * mid * tmp + 1;
        break;
      }
      tmp *= 2;
      if (d >= k) break;
    }
    if (d >= k) {
      r2 = mid * 2;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  cout << max(r1, r2);
}
