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 n, k, m, dp[100][2];
string s, ans1;
long long check(string s, string lim) {
  int len = s.length();
  long long ans = 0;
  if (s[len - 1] == '0') {
    for (int i = len; i < m; i++) {
      ans += 2 * (1LL << (i - len));
    }
    s = s.substr(0, len - 1);
    len--;
  } else {
    for (int i = len; i < m; i++) {
      ans += (1LL << (i - len));
    }
  }
  int kt = 0;
  for (int i = 0; i < len; i++) {
    if (s[i] != lim[i]) {
      if (s[i] < lim[i]) {
        kt = 1;
      } else {
        kt = 2;
      }
      break;
    }
  }
  if (kt == 0) {
    memset(dp, 0, sizeof dp);
    dp[len][1] = 1;
    for (int i = len + 1; i <= m; i++) {
      for (int cs = 0; cs <= 1; cs++) {
        if (cs == lim[i - 1] - 48) {
          dp[i][1] += dp[i - 1][1];
        }
        if (cs < lim[i - 1] - 48) {
          dp[i][0] += dp[i - 1][1];
        }
        dp[i][0] += dp[i - 1][0];
      }
    }
    ans += dp[m][0] + dp[m][1];
  } else {
    if (kt == 1) {
      ans += (1LL << (m - len));
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  long long tmp1 = n, ans = 0;
  while (tmp1 > 0) {
    s = (char)(tmp1 % 2 + 48) + s;
    tmp1 /= 2;
  }
  m = s.length();
  for (int len = 1; len <= m; len++) {
    string tmp;
    while (tmp.length() < len) tmp += "0";
    for (int i = 0; i < len; i++) {
      tmp[i] = '1';
      if (check(tmp, s) < k) {
        tmp[i] = '0';
      }
    }
    if (check(tmp, s) >= k) {
      long long tmp1 = 0;
      for (int i = 0; i < len; i++) {
        tmp1 = 2 * tmp1 + tmp[i] - 48;
      }
      ans = max(ans, tmp1);
    }
  }
  cout << ans;
}
