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 target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long int inf = 1e18 + 9;
long long int ans = 0;
bool check1(long long int mid, long long int ile, long long int k,
            long long int n) {
  long long int ans = 2;
  if (mid + 1 > n) {
    if (1 >= k) return true;
    return false;
  }
  long long int biore = 2;
  for (long long int i = 1; i <= ile; i++) {
    mid *= 2;
    biore *= 2;
    if (mid > n) break;
    if (mid + biore - 1 <= n) {
      ans += biore;
    } else {
      ans += min(n - mid + 1, biore);
      break;
    }
  }
  if (ans >= k)
    return true;
  else
    return false;
}
bool check2(long long int mid, long long int ile, long long int k,
            long long int n) {
  long long int ans = 1;
  long long int biore = 1;
  for (long long int i = 1; i <= ile; i++) {
    mid *= 2;
    biore *= 2;
    if (mid > n) break;
    if (mid + biore - 1 <= n) {
      ans += biore;
    } else {
      ans += min(n - mid + 1, biore);
      break;
    }
  }
  if (ans >= k)
    return true;
  else
    return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, k;
  cin >> n >> k;
  if (n == k) {
    cout << "1\n";
  } else if (k == n - 1) {
    cout << "2\n";
  } else if (k == 1)
    cout << n << "\n";
  else {
    long long int wskaznik = -1;
    for (int i = 1; i <= 60; i++) {
      if ((1LL << i) - 1 >= k) {
        wskaznik = i;
        break;
      }
    }
    long long int ip1 = 0, ik1 = (n + 1) / 2 + 1;
    while (ip1 + 1 < ik1) {
      long long int mid = ((ip1 + ik1) / 2) * 2;
      if (check1(mid, wskaznik, k, n)) {
        ip1 = mid / 2;
      } else
        ik1 = mid / 2;
    }
    long long int ip2 = 0, ik2 = (n + 1) / 2 + 1;
    while (ip2 + 1 < ik2) {
      long long int mid = ((ip2 + ik2) / 2) * 2 + 1;
      if (check2(mid, wskaznik, k, n)) {
        ip2 = (mid - 1) / 2;
      } else
        ik2 = (mid - 1) / 2;
    }
    ans = max(2LL * ip1, 2LL * ip2 + 1);
    cout << ans << "\n";
  }
  return 0;
}
