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;
unsigned long long int N, k;
string tgt;
string toStr(unsigned long long int x) {
  string res = "";
  while (x) {
    res = (char)(x % 2 + '0') + res;
    x /= 2;
  }
  return res;
}
unsigned long long int f(string str) {
  string cmp = tgt.substr(0, str.length());
  unsigned long long int res;
  if (cmp == str) {
    res = (1ULL << (tgt.length() - str.length())) - 1;
    for (unsigned long long int i = str.length(); i < tgt.length(); i++) {
      if (tgt[i] == '1') {
        res += (1ULL << (tgt.length() - i - 1));
      }
    }
    res++;
    if (str[str.length() - 1] == '0') {
      str[str.length() - 1] = '1';
      res += f(str);
    }
  } else if (cmp > str) {
    res = (1ULL << (tgt.length() - str.length() + 1)) - 1;
    if (str[str.length() - 1] == '0') {
      str[str.length() - 1] = '1';
      res += f(str);
    }
  } else {
    res = (1ULL << (tgt.length() - str.length())) - 1;
    if (str[str.length() - 1] == '0') {
      str[str.length() - 1] = '1';
      res += f(str);
    }
  }
  return res;
}
unsigned long long int getPath(unsigned long long int x) {
  if (x > N)
    return 0;
  else
    return f(toStr(x));
}
signed main() {
  cin >> N >> k;
  tgt = toStr(N);
  unsigned long long int l = 1, r = (N + 10) / 2, m, a, b;
  while (l + 1 < r) {
    m = (l + r) / 2;
    if (getPath(2 * m) >= k) {
      l = m;
    } else
      r = m;
  }
  a = 2 * l;
  l = 0, r = (N + 10) / 2;
  while (l + 1 < r) {
    m = (l + r) / 2;
    if (getPath(2 * m + 1) >= k) {
      l = m;
    } else
      r = m;
  }
  b = 2 * l + 1;
  if (getPath(a) < k)
    cout << b << endl;
  else
    cout << max(a, b) << endl;
}
