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;
void solve();
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed;
  cout.precision(20);
  solve();
  return 0;
}
template <typename T>
void chmin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
void chmax(T& a, T b) {
  if (a < b) a = b;
}
template <typename T>
void MACRO_rdv2_Init(int n, T& t) {
  t.resize(n);
}
template <typename First, typename... Rest>
void MACRO_rdv2_Init(int n, First& first, Rest&... rest) {
  first.resize(n);
  MACRO_rdv2_Init(n, rest...);
}
template <typename T>
void MACRO_rdv2_Scan(int p, T& t) {
  std::cin >> t[p];
}
template <typename First, typename... Rest>
void MACRO_rdv2_Scan(int p, First& first, Rest&... rest) {
  std::cin >> first[p];
  MACRO_rdv2_Scan(p, rest...);
}
template <typename T>
void wrv(const vector<T>& v) {
  for (int(__ii) = (0); (__ii) < (((int)v.size())); ++(__ii)) {
    if (__ii) cout << ' ';
    cout << v[__ii];
  }
  cout << '\n';
}
template <typename T>
void wrm(const vector<vector<T>>& v) {
  for (int(__ii) = (0); (__ii) < (((int)v.size())); ++(__ii)) {
    for (int(__jj) = (0); (__jj) < (v[__ii].size()); ++(__jj)) {
      if (__jj) cout << ' ';
      cout << v[__ii][__jj];
    }
    cout << '\n';
  }
}
template <typename T>
void sc(T& x) {
  cin >> x;
}
template <typename Head, typename... Tail>
void sc(Head& head, Tail&... tail) {
  cin >> head;
  sc(tail...);
}
template <typename T>
void wr(const T& x) {
  cout << x << '\n';
}
template <typename Head, typename... Tail>
void wr(const Head& head, const Tail&... tail) {
  cout << head << ' ';
  wr(tail...);
}
template <typename T>
void wrf(const T& x) {
  cout << x << endl;
}
template <typename Head, typename... Tail>
void wrf(const Head& head, const Tail&... tail) {
  cout << head << ' ';
  wrf(tail...);
}
template <typename T>
void debug_out(const T& x) {
  cerr << x << '\n';
}
template <typename Head, typename... Tail>
void debug_out(const Head& head, const Tail&... tail) {
  cerr << head << ' ';
  debug_out(tail...);
}
template <typename... T>
void err(const T&... cod) {
  wr(cod...);
  exit(0);
}
long long get(long long x, long long n) {
  long long len = 1 + (x % 2 == 0), cnt = 0;
  while (x + len - 1 < n) {
    cnt += len;
    x *= 2;
    len *= 2;
  }
  cnt += max(0LL, n - x + 1);
  return cnt;
}
void solve() {
  long long n, k;
  sc(n, k);
  long long ans = 1;
  long long low = -1, high = n + 1;
  while (low + 1 < high) {
    long long mid = (low + high) / 2;
    if (get(2 * mid, n) >= k)
      low = mid;
    else
      high = mid;
  }
  chmax(ans, 2 * low);
  low = -1, high = n + 1;
  while (low + 1 < high) {
    long long mid = (low + high) / 2;
    if (get(2 * mid + 1, n) >= k)
      low = mid;
    else
      high = mid;
  }
  chmax(ans, 2 * low + 1);
  wr(ans);
}
