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;
long long modulo(long long base, long long exponent, long long modulus);
long long choose(long long n, long long k);
long long inverse(long long a, long long m);
void build();
void fileio();
long long ncr(long long n, long long r);
const int nax = 1e6 + 10;
const int LG = log2(nax) + 1;
vector<long long> fact(nax);
long long get(long long x, long long n) {
  long long ans = 0;
  for (long long j = 0; j <= 70; j++) {
    long long t = x << j;
    if (t > n) break;
    long long one = n - t + 1;
    ans += min(one, 1LL << j);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long low = 1, high = (n + 1) / 2;
  long long ans = 1;
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (get(2 * mid - 1, n) >= k) {
      ans = 2 * mid - 1;
      low = mid + 1;
    } else
      high = mid - 1;
  }
  low = 1, high = n / 2;
  long long ans2 = 0;
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (get(2 * mid + 1, n) + get(2 * mid, n) >= k) {
      ans2 = 2 * mid;
      low = mid + 1;
    } else
      high = mid - 1;
  }
  cout << max(ans, ans2) << '\n';
  return 0;
}
long long ncr(long long n, long long r) {
  if (r > n || r < 0 || n < 0) return 0;
  long long ans = fact[n];
  long long temp = (fact[n - r] * fact[r]) % 1000000007;
  ans = (ans * inverse(temp, 1000000007)) % 1000000007;
  return ans;
}
void fileio() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void build() {
  fact[0] = 1;
  for (long long i = 1; i < nax; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long modulo(long long base, long long exponent, long long modulus) {
  if (modulus == 1) return 0;
  long long result = 1;
  base = base % modulus;
  while (exponent > 0) {
    if (exponent % 2 == 1) {
      result = (result * base) % modulus;
    }
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
void EE(long long a, long long b, long long &co1, long long &co2) {
  if (a % b == 0) {
    co1 = 0;
    co2 = 1;
    return;
  }
  EE(b, a % b, co1, co2);
  long long temp = co1;
  co1 = co2;
  co2 = temp - co2 * (a / b);
}
long long inverse(long long a, long long m) {
  long long x, y;
  EE(a, m, x, y);
  if (x < 0) x += m;
  return x;
}
