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;
unsigned long long pw[64];
bool check(unsigned long long x, unsigned long long n, unsigned long long k) {
  if (!x) return 1;
  string N;
  while (n) N = char(n % 2 + '0') + N, n /= 2;
  string X;
  while (x) X = char(x % 2 + '0') + X, x /= 2;
  unsigned long long ans = 0;
  if (X.size() < N.size()) {
    for (int i = X.size(); i < (int)N.size(); ++i)
      ans += pw[i - (int)X.size() + (X.back() == '0')];
  }
  int xsiz = X.size();
  while (X.size() < N.size()) X += '0';
  if (X <= N) {
    vector<unsigned long long> dp[2];
    dp[0].resize(N.size() + 1);
    dp[1].resize(N.size() + 1);
    dp[0][0] = 1;
    for (int i = 0; i < N.size(); ++i) {
      for (int f = 0; f < 2; ++f) {
        dp[f | (X[i] < N[i])][i + 1] += dp[f][i];
        if (i >= xsiz - 1 && X[i] == '0' && (f || X[i] < N[i]))
          dp[f][i + 1] += dp[f][i];
      }
    }
    ans += dp[0][N.size()] + dp[1][N.size()];
  }
  return ans >= k;
}
int main() {
  pw[0] = 1;
  for (int i = 1; i < 64; ++i) pw[i] = pw[i - 1] * 2;
  unsigned long long n, k;
  scanf("%llu %llu", &n, &k);
  unsigned long long lb = 0, rb = n / 2;
  while (lb < rb) {
    unsigned long long md = (lb + rb + 1) >> 1;
    if (check(md * 2, n, k))
      lb = md;
    else
      rb = md - 1;
  }
  unsigned long long ans = lb * 2;
  lb = 0;
  rb = (n - 1) / 2;
  while (lb < rb) {
    unsigned long long md = (lb + rb + 1) >> 1;
    if (check(md * 2 + 1, n, k))
      lb = md;
    else
      rb = md - 1;
  }
  ans = max(ans, lb * 2 + 1);
  printf("%llu\n", ans);
  return 0;
}
