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;
bool check(long long n, long long pos) { return n & (1LL << pos); }
long long Set(long long n, long long pos) { return n = n | (1LL << pos); }
long long freq[1000];
vector<long long> app[1000];
void f(long long n, long long i) {
  freq[i]++;
  app[i].push_back(n);
  if (i == 1) {
    cout << 1 << endl;
    return;
  }
  cout << i << " ";
  if (i % 2)
    f(n, i - 1);
  else
    f(n, i / 2);
}
void fndpattern() {
  for (long long i = 1; i <= 1000; i++) {
    cout << i << ": ";
    f(i, i);
    for (long long j = 1; j <= i; j++) {
      if (j % 2) continue;
      cout << freq[j] << " ";
    }
    cout << endl;
    for (long long j = 1; j <= i; j++) {
      if (j % 2 == 0) continue;
      cout << freq[j] << " ";
    }
    cout << endl;
  }
  for (long long i = 1; i <= 100; i++) {
    cout << i << "!:: ";
    for (auto it : app[i]) cout << it << " ";
    cout << endl;
  }
}
long long pw2[100];
long long cnteven(long long n, long long p) {
  long long x = 0;
  long long i = 0;
  long long ret = 0;
  while ((pw2[i] * p) <= n) {
    ret += (pw2[i] * 2);
    i++;
  }
  i--;
  ret -= (pw2[i] * 2);
  x = pw2[i] * p;
  ret += min(pw2[i] * 2, n - x + 1);
  return ret;
}
long long cntodd(long long n, long long p) {
  long long x = 0;
  long long i = 0;
  long long ret = 0;
  while ((pw2[i] * p) <= n) {
    ret += (pw2[i]);
    i++;
  }
  i--;
  ret -= (pw2[i]);
  x = pw2[i] * p;
  ret += min(pw2[i], n - x + 1);
  return ret;
}
long long solveeven(long long n, long long k) {
  long long lo = 1;
  long long hi = (n / 2);
  long long ret = 0;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    if (cnteven(n, mid * 2) >= k) {
      ret = mid * 2;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return ret;
}
long long solveodd(long long n, long long k) {
  long long lo = 0;
  long long hi = (n / 2);
  long long ret = 0;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    if ((mid * 2) + 1 > n) {
      hi = mid - 1;
    } else if (cntodd(n, (mid * 2) + 1) >= k) {
      ret = (mid * 2) + 1;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return ret;
}
int main() {
  long long i, j, k, l, m, n, o, r, q;
  long long testcase;
  long long input, flag, tag, ans;
  pw2[0] = 1;
  for (i = 1; i <= 61; i++) {
    pw2[i] = 2 * pw2[i - 1];
  }
  while (cin >> n >> k) {
    ans = max(solveeven(n, k), solveodd(n, k));
    printf("%lld\n", ans);
  }
}
