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;
const int MAXN = 64 + 20;
long long n, k;
long long d[MAXN][3];
long long ln(long long a) {
  long long l = 0;
  while (a) {
    a /= 2;
    l++;
  }
  return l;
}
long long getnum(long long l) {
  long long alen = ln(l), blen = ln(n);
  memset(d, 0, sizeof(d));
  if ((n >> (blen - alen)) > (l))
    d[alen][0] = 1;
  else if ((n >> (blen - alen)) == (l))
    d[alen][1] = 1;
  else if ((n >> (blen - alen)) < (l))
    d[alen][2] = 1;
  if ((l & 1) == 0 && (l + 1) <= n) {
    if ((n >> (blen - alen)) > (l + 1))
      d[alen][0] += 1;
    else if ((n >> (blen - alen)) == (l + 1))
      d[alen][1] += 1;
    else if ((n >> (blen - alen)) < (l + 1))
      d[alen][2] += 1;
  }
  long long ans = d[alen][0] + d[alen][1] + d[alen][2];
  for (long long i = alen; i < blen; ++i) {
    if ((n >> (blen - i - 1)) & 1) {
      d[i + 1][0] = d[i][0] * 2 + d[i][1];
      d[i + 1][2] = d[i][2] * 2;
      d[i + 1][1] = d[i][1];
    } else {
      d[i + 1][0] = d[i][0] * 2;
      d[i + 1][2] = d[i][2] * 2 + d[i][1];
      d[i + 1][1] = d[i][1];
    }
    if (i != blen - 1) {
      for (long long j = 0; j < 3; ++j) {
        ans += d[i + 1][j];
      }
    } else {
      for (long long j = 0; j < 2; ++j) {
        ans += d[i + 1][j];
      }
    }
  }
  return ans;
}
int main() {
  cin >> n >> k;
  if (n == k) {
    puts("1");
    return 0;
  }
  if (k == 1) {
    cout << n;
    return 0;
  }
  long long l = 1, r = n / 2;
  while (l < r) {
    long long mid = (l + r + 1) / 2;
    if (getnum(mid * 2) >= k)
      l = mid;
    else
      r = mid - 1;
  }
  long long ans = l * 2;
  l = 1, r = (n + 1) / 2;
  while (l < r) {
    long long mid = (l + r + 1) / 2;
    if (getnum(mid * 2 - 1) >= k)
      l = mid;
    else
      r = mid - 1;
  }
  cout << max(ans, l * 2 - 1) << endl;
}
