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;
ostream &operator<<(ostream &a, const vector<long long> &b) {
  for (auto k : b) cout << k << " ";
  return a;
}
const long long INF = 8e18;
const long long p = 179;
long long n, k;
const long long LOG = 60;
long long sz = 0;
bitset<LOG> a;
long long count(long long k) {
  if (k > n) {
    return 0;
  } else if (k == n) {
    return 1;
  }
  bitset<LOG> cur(k);
  long long h = 0;
  for (long long i = 0; i < LOG; i++) {
    if (cur[i]) h = i;
  }
  if (sz == h) {
    return 2 - (k & 1);
  }
  long long ans = 0;
  if (k & 1) {
    ans = 1;
    for (long long i = h + 1; i < sz; i++) {
      ans += (1ll << (i - h));
    }
    if ((k << (sz - h)) > n) {
      return ans;
    }
    long long l = 0, r = (1ll << LOG), mid;
    while (r - l > 1) {
      mid = (l + r) / 2;
      if (mid + (k << (sz - h)) <= n)
        l = mid;
      else
        r = mid;
    }
    ans += min(l + 1, (1ll << (sz - h)));
  } else {
    ans = 2;
    for (long long i = h + 1; i < sz; i++) {
      ans += (1ll << (i - h + 1));
    }
    if ((k << (sz - h)) > n) {
      return ans;
    }
    long long l = 0, r = (1ll << LOG), mid;
    while (r - l > 1) {
      mid = (l + r) / 2;
      if (mid + (k << (sz - h)) <= n)
        l = mid;
      else
        r = mid;
    }
    ans += min(l + 1, (1ll << (sz - h + 1)));
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  a = bitset<LOG>(n);
  for (long long i = 0; i < LOG; i++) {
    if (a[i]) sz = i;
  }
  long long l = 0, r = 6e17, mid;
  long long ans = 1;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (count(2 * mid) >= k)
      l = mid;
    else
      r = mid;
  }
  ans = max(ans, 2 * l);
  l = 0, r = 6e17;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (count(2 * mid + 1) >= k)
      l = mid;
    else
      r = mid;
  }
  ans = max(ans, 2 * l + 1);
  cout << ans;
}
