#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long mul[64];
long long get(long long x) {
  int lena;
  for (int i = 1; i < 63; i++) {
    if (mul[i] > n) {
      lena = i;
      break;
    }
  }
  int lenb;
  for (int i = 1; i < 63; i++) {
    if (mul[i] > x) {
      lenb = i;
      break;
    }
  }
  int lenx = lena - lenb;
  long long sa = n / mul[lenx];
  if (sa > x)
    return mul[lenx + 1] - 1;
  else if (sa == x)
    return mul[lenx] + n % mul[lenx];
  else
    return mul[lenx] - 1;
}
long long check(long long x) {
  if (x % 2 == 0 && x < n)
    return get(x) + get(x + 1);
  else
    return get(x);
}
int main() {
  scanf("%lld %lld", &n, &k);
  long long l = 0, r = n / 2;
  mul[0] = 1;
  for (int i = 1; i < 63; i++) mul[i] = mul[i - 1] * 2;
  while (r - l > 1) {
    long long mid = (r + l) >> 1;
    if (check(mid * 2) >= k)
      l = mid;
    else
      r = mid;
  }
  if (check(r * 2 + 1) >= k)
    printf("%lld\n", r * 2 + 1);
  else if (check(r * 2) >= k)
    printf("%lld\n", r * 2);
  else if (check(l * 2 + 1) >= k)
    printf("%lld\n", l * 2 + 1);
  else
    printf("%lld\n", l * 2);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long mul[64];
long long get(long long x) {
  int lena;
  for (int i = 1; i < 63; i++) {
    if (mul[i] > n) {
      lena = i;
      break;
    }
  }
  int lenb;
  for (int i = 1; i < 63; i++) {
    if (mul[i] > x) {
      lenb = i;
      break;
    }
  }
  int lenx = lena - lenb;
  long long sa = n / mul[lenx];
  if (sa > x)
    return mul[lenx + 1] - 1;
  else if (sa == x)
    return mul[lenx] + n % mul[lenx];
  else
    return mul[lenx] - 1;
}
long long check(long long x) {
  if (x % 2 == 0 && x < n)
    return get(x) + get(x + 1);
  else
    return get(x);
}
int main() {
  scanf("%lld %lld", &n, &k);
  long long l = 0, r = n / 2;
  mul[0] = 1;
  for (int i = 1; i < 63; i++) mul[i] = mul[i - 1] * 2;
  while (r - l > 1) {
    long long mid = (r + l) >> 1;
    if (check(mid * 2) >= k)
      l = mid;
    else
      r = mid;
  }
  if (check(r * 2 + 1) >= k)
    printf("%lld\n", r * 2 + 1);
  else if (check(r * 2) >= k)
    printf("%lld\n", r * 2);
  else if (check(l * 2 + 1) >= k)
    printf("%lld\n", l * 2 + 1);
  else
    printf("%lld\n", l * 2);
  return 0;
}

```