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 l, r, n, k, len, it, m, dp[107];
string s;
string get_s(long long n) {
  string s = "";
  while (n > 0) {
    s += (n % 2 + '0');
    n /= 2;
  }
  return s;
}
long long f(long long a) {
  string r = get_s(a);
  reverse(r.begin(), r.end());
  long long l = r.length();
  long long sl = l;
  long long it = 1;
  long long cnt = 0;
  if (sl == len) {
    long long ret = 0;
    if (r <= s) ret = 1;
    if (a % 2 == 0) ret += f(a + 1);
    return ret;
  }
  while (l < len) {
    cnt += it;
    it *= 2;
    r += '0';
    l++;
  }
  long long flag = 0;
  for (int i = 0; i < sl; i++) {
    if (r[i] > s[i]) {
      return cnt;
    }
    if (r[i] < s[i]) {
      flag = 1;
      break;
    }
  }
  if (a % 2 == 0) cnt += f(a + 1);
  if (flag == 0)
    return cnt + dp[sl];
  else
    return cnt + it;
}
int main() {
  cin >> n >> k;
  s = get_s(n);
  len = s.length();
  reverse(s.begin(), s.end());
  if (s[len - 1] == '0')
    dp[len - 1] = 1;
  else
    dp[len - 1] = 2;
  it = 2;
  for (int i = len - 2; i >= 0; i--) {
    if (s[i] == '0')
      dp[i] = dp[i + 1];
    else
      dp[i] = dp[i + 1] + it;
    it *= 2;
  }
  l = 1;
  r = n;
  if (f(r) >= k) {
    cout << r;
    return 0;
  }
  r /= 2;
  while (r - l > 2) {
    m = (l + r - 1) / 2;
    if (f(2 * m) < k)
      r = m - 1;
    else
      l = m;
  }
  if (f(2 * l) < k)
    cout << 1;
  else {
    for (long long i = l + 1; i <= r + 1; i++) {
      if (f(2 * i) < k) {
        l = i - 1;
        break;
      }
    }
    if (f(2 * l + 1) >= k)
      cout << 2 * l + 1;
    else
      cout << 2 * l;
  }
  return 0;
}
