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>
#pragma GCC optimize("O3")
using namespace std;
string itos(int n) {
  stringstream ss;
  ss << n;
  string second = ss.str();
  return second;
}
long long ppow(long long x, long long y, long long mod) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return res;
}
int main() {
  cout << fixed << setprecision(10);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long l = 0, r = n;
  while (l < r) {
    long long m = (l + r + 1) / 2;
    long long x = m * 2;
    long long y = x + 1;
    long long cur = 0;
    while (x <= n) {
      cur += min(y, n) - x + 1;
      x = x * 2;
      y = y * 2 + 1;
    }
    if (cur >= k) {
      l = m;
    } else {
      r = m - 1;
    }
  }
  long long ans1 = l + l;
  l = 0, r = n;
  while (l < r) {
    long long m = (l + r + 1) / 2;
    long long x = m + m + 1;
    long long y = m + m + 1;
    long long cur = 0;
    while (x <= n) {
      cur += min(y, n) - x + 1;
      x = x + x;
      y = y + y + 1;
    }
    if (cur >= k) {
      l = m;
    } else {
      r = m - 1;
    }
  }
  long long ans2 = l + l + 1;
  cout << max(ans1, ans2);
}
