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;
inline long long read() {
  long long t = 1, s = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * t;
}
long long n, k;
bool judge(long long x) {
  if (x & 1) {
    long long ans = 1;
    for (long long i = 2; x * i <= n; i *= 2) {
      long long number = n - x * i;
      if (number < i) {
        ans += number + 1;
        break;
      } else
        ans += i;
    }
    return ans >= k;
  } else {
    long long ans = 0;
    for (long long i = 1; x * i <= n; i *= 2) {
      long long number = n - x * i;
      if (number < 2 * i) {
        ans += number + 1;
        break;
      } else
        ans += 2 * i;
    }
    return ans >= k;
  }
}
signed main(void) {
  n = read(), k = read();
  long long l = 1, r = (n + 1) / 2, ans_fir = 1;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (judge(mid * 2 - 1)) {
      ans_fir = mid * 2 - 1;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  l = 1, r = n / 2;
  long long ans_sec = 0;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (judge(mid * 2)) {
      ans_sec = mid * 2;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  printf("%lld\n", max(ans_sec, ans_fir));
  return 0;
}
