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>
struct node {
  long long ln, lv;
  long long rn, rv;
  long long n[6], v[6];
} a[100];
int tot = 0;
long long n, k, b;
int main() {
  int i, j, check;
  long long cmd;
  scanf("%lld%lld", &n, &k), b = 1;
  for (i = 1;; i++) {
    if (n == 0) break;
    a[i].rv = n, a[i].rn = a[i - 1].ln * (a[i - 1].lv % 2) + 1;
    for (j = 1; j <= 4; j++)
      if (a[i].rv * 2 == a[i - 1].v[j]) break;
    a[i].rn += a[i - 1].n[j];
    a[i].lv = n / 2 + 1;
    cmd = a[i].rv - a[i].lv + 1;
    if (cmd >= 4) {
      a[i].n[1] = a[i].rn, a[i].v[1] = a[i].rv;
      for (j = 2; j <= 4; j++) a[i].v[j] = a[i].v[j - 1] - 1;
      for (j = 1; j <= 4; j++)
        if (a[i].v[2] * 2 == a[i - 1].v[j]) break;
      a[i].n[2] = a[i - 1].n[j] + a[i].n[1] * (a[i].v[1] % 2) + 1;
      a[i].n[3] = b + b * ((a[i].v[3] - 1) % 2);
      a[i].n[4] = b + b * ((a[i].v[4] - 1) % 2);
      a[i].ln = b + b * ((a[i].lv - 1) % 2);
    } else if (cmd == 3) {
      a[i].n[1] = a[i].rn, a[i].v[1] = a[i].rv;
      for (j = 2; j <= 3; j++) a[i].v[j] = a[i].v[j - 1] - 1;
      a[i].lv = a[i].v[4] = a[i].v[3];
      for (j = 1; j <= 4; j++)
        if (a[i].v[2] * 2 == a[i - 1].v[j]) break;
      a[i].n[2] = a[i - 1].n[j] + a[i].n[1] * (a[i].v[1] % 2) + 1;
      a[i].n[3] = b + b * ((a[i].v[3] - 1) % 2);
      a[i].ln = a[i].n[4] = a[i].n[3];
    } else if (cmd == 2) {
      a[i].n[1] = a[i].rn, a[i].v[1] = a[i].rv;
      for (j = 2; j <= 4; j++) a[i].v[j] = a[i].rv - 1;
      for (j = 1; j <= 4; j++)
        if (a[i].v[2] * 2 == a[i - 1].v[j]) break;
      a[i].n[2] = a[i - 1].n[j] + a[i].n[1] * (a[i].v[1] % 2) + 1;
      for (j = 3; j <= 4; j++) a[i].n[j] = a[i].n[2];
      a[i].ln = a[i].n[2];
    } else if (cmd == 1) {
      for (j = 1; j <= 4; j++) a[i].n[j] = a[i].rn, a[i].v[j] = a[i].rv;
      a[i].ln = a[i].rn;
    }
    n /= 2, tot++;
    b = b * 2 + 1;
  }
  check = 0;
  for (i = 1; i <= tot; i++) {
    for (j = 1; j <= 4; j++)
      if (a[i].n[j] >= k) {
        check = 1;
        break;
      }
    if (check) break;
  }
  printf("%lld\n", a[i].v[j]);
  return 0;
}
