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 LOG = 60;
int bit(long long mask, int k) { return (mask >> k) & 1; }
void printm(long long mask) {
  for (int k = 0; k < 5; ++k) printf("%d", bit(mask, k));
  printf("\n");
}
long long pw(int k) {
  if (k < 0) return 0;
  return 1LL << k;
}
long long get2(long long x, long long n, int k) {
  if (x > n) return 0;
  int pos = LOG;
  while (pos >= 0 && bit(n, pos) == bit(x, pos)) --pos;
  long long res;
  if (pos >= k)
    res = 1LL << k;
  else
    res = n - x + 1;
  return res;
}
long long get(long long x, long long n, int k) {
  long long res = 0;
  for (int i = 0; i <= LOG - k + 1; ++i) res += get2(x << i, n, i);
  if (x % 2 == 0) {
    x ^= 1;
    for (int i = 0; i <= LOG - k + 1; ++i) res += get2(x << i, n, i);
    x ^= 1;
  }
  return res;
}
int main() {
  long long n, k;
  scanf("%lld %lld", &n, &k);
  long long ans = 0;
  for (int i = LOG; i >= 0; --i) {
    long long cur = 1LL << i;
    if (get(cur, n, i + 1) < k) continue;
    for (int j = i - 1; j >= 0; --j) {
      cur ^= 1LL << j;
      if (get(cur, n, i + 1) < k) cur ^= 1LL << j;
    }
    printf("%lld\n", cur);
    return 0;
  }
}
