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 limit = 998244353;
bool cmp(long long a, long long b) { return a > b; }
long long myceil(long long a, long long b) { return a / b + bool(a % b); }
long long mylog(long long num, long long base) {
  long long ans = 0;
  while (num) {
    ans++;
    num /= base;
  }
  return ans - 1;
}
long long Pow(long long a, long long b) {
  long long ans = 1, mul = a;
  while (b) {
    if (b & 1) ans = (ans * mul) % limit;
    mul = (mul * mul) % limit;
    b /= 2;
  }
  return ans;
}
int main() {
  int t = 1;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long nn = n, mm = m;
    long long ans1, ans2;
    long long tmp = 1;
    long long cnt1 = 0, cnt2 = 0;
    while (m > tmp) {
      m -= tmp;
      tmp *= 2;
      cnt1++;
    }
    ans1 = n - (m - 1);
    ans1 /= pow(2, cnt1);
    tmp = 2;
    m = mm;
    while (m > tmp) {
      m -= tmp;
      tmp *= 2;
      cnt2++;
    }
    ans2 = n - (m - 1);
    ans2 /= pow(2, cnt2);
    long long ans;
    if (ans1 >= ans2 && ans1 % 2) {
      ans = ans1;
    } else if (ans1 >= ans2)
      ans = max(ans - 1, ans2);
    else if (ans2 >= ans1 && ans2 % 2 == 0)
      ans = ans2;
    else
      ans = max(ans1, ans2 - 1);
    cout << ans << "\n";
  }
}
