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 MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFL = 0x3f3f3f3f3f3f3f3f;
long long n, K;
long long dp[70][2];
int CNT(long long n) {
  int res = 0;
  while (n) {
    res++;
    n >>= 1;
  }
  return res;
}
long long C(long long d) {
  int S = CNT(n);
  int D = CNT(d);
  long long res = 0;
  for (int I = S; I >= D; I--) {
    memset(dp, 0, sizeof(dp));
    if (I == S)
      dp[0][0] = 1;
    else
      dp[0][1] = 1;
    for (int i = 0; i < (I); i++)
      for (int j = 0; j < (2); j++) {
        if (dp[i][j] == 0) continue;
        int lim = (j ? 1 : (n >> (I - i - 1) & 1));
        for (int k = 0; k <= lim; k++) {
          if (i < D) {
            if (i == D - 1 && d % 2 == 0)
              ;
            else {
              int w = d >> (D - i - 1) & 1;
              if (k != w) continue;
            }
          }
          if (i == 0 && !k) continue;
          dp[i + 1][j || k < lim] += dp[i][j];
        }
      }
    res += dp[I][0] + dp[I][1];
  }
  return res;
}
int main() {
  cin >> n >> K;
  long long ok = 0, ng = n / 2 + 1;
  while (ng - ok > 1) {
    long long t = (ng + ok) / 2;
    if (C(2 * t) >= K)
      ok = t;
    else
      ng = t;
  }
  long long ok2 = 0, ng2 = (n - 1) / 2 + 1;
  while (ng2 - ok2 > 1) {
    long long t = (ng2 + ok2) / 2;
    if (C(2 * t + 1) >= K)
      ok2 = t;
    else
      ng2 = t;
  }
  cout << max(ok * 2, ok2 * 2 + 1) << endl;
}
