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(2)
using namespace std;
const int mod = 1e9 + 7;
template <typename T>
void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return a / gcd(a, b) * b; };
long long ksm(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = (ans * a) % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans % mod;
}
long long n, k;
long long get_len(long long x) {
  long long len = 0;
  while (x) {
    len++;
    x >>= 1;
  }
  return len;
}
long long cal2(long long x) {
  long long pos2 = 1;
  long long res = 0;
  while (x <= n) {
    if (((n & (~(pos2 - 1)))) == x)
      res += n - x + 1;
    else
      res += pos2;
    pos2 <<= 1;
    x <<= 1;
  }
  return res;
}
long long now = 0;
long long cal(long long pos) {
  if (!(pos & 1)) {
    long long res1 = cal2(pos);
    long long res2 = cal2(pos | 1);
    return res1 + res2;
  } else {
    long long res = cal2(pos);
    return res;
  }
}
bool check(long long x) {
  long long cnt = cal(x);
  return cnt >= k;
}
long long get_ans1(long long seed) {
  long long l = 1, r = seed;
  while (l <= r) {
    long long m = (l + r) >> 1;
    if (check(2 * m + 1)) {
      l = m + 1;
    } else {
      r = m - 1;
    }
  }
  return 2 * (l - 1) + 1;
}
inline long long get_ans2(long long seed) {
  long long l = 1, r = seed;
  while (l <= r) {
    long long m = (l + r) >> 1;
    if (check(2 * m)) {
      l = m + 1;
    } else {
      r = m - 1;
    }
  }
  return 2 * (l - 1);
}
inline long long solve() { return max(get_ans1(n >> 1), get_ans2(n >> 1)); }
signed main() {
  read(n), read(k);
  write(solve());
  return 0;
}
