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;
char inline nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <typename T>
bool rd(T& v) {
  static char ch;
  while (ch != EOF && !isdigit(ch)) ch = nc();
  if (ch == EOF) return false;
  for (v = 0; isdigit(ch); ch = nc()) v = v * 10 + ch - '0';
  return true;
}
template <typename T>
void o(T p) {
  static int stk[70], tp;
  if (p == 0) {
    putchar('0');
    return;
  }
  if (p < 0) {
    p = -p;
    putchar('-');
  }
  while (p) stk[++tp] = p % 10, p /= 10;
  while (tp) putchar(stk[tp--] + '0');
}
const int maxn = 1e5 + 5;
const int maxm = 4e5 + 5;
const int inf = 1e9;
long long n, k;
long long count_odd(long long x) {
  long long ans = 0;
  long long cnt = 1;
  while (x <= n) {
    ans += min(n - x + 1, cnt);
    cnt <<= 1;
    x <<= 1;
  }
  return ans;
}
long long count_even(long long x) {
  long long ans = 0;
  long long cnt = 2;
  while (x <= n) {
    ans += min(n - x + 1, cnt);
    cnt <<= 1;
    x <<= 1;
  }
  return ans;
}
long long solve_odd() {
  long long l = 0, r = n / 2 + 1;
  while (l < r) {
    long long mid = (l + r + 1) >> 1;
    if (count_odd(2 * mid + 1) >= k)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
long long solve_even() {
  long long l = 0, r = n / 2 + 1;
  while (l < r) {
    long long mid = (l + r + 1) >> 1;
    if (count_even(2 * mid) >= k)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
int main() {
  cin >> n >> k;
  cout << max(2 * solve_odd() + 1, 2 * solve_even()) << endl;
}
