Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains n sweet candies from the good ol' bakery, each labeled from 1 to n corresponding to its tastiness. No two candies have the same tastiness.

The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!

A xor-sum of a sequence of integers a1, a2, ..., am is defined as the bitwise XOR of all its elements: <image>, here <image> denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)

Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than k candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.

Input

The sole string contains two integers n and k (1 ≤ k ≤ n ≤ 1018).

Output

Output one number — the largest possible xor-sum.

Examples

Input

4 3


Output

7


Input

6 6


Output

7

Note

In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.

In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
template <typename X>
inline X abs(const X& a) {
  return a < 0 ? -a : a;
}
template <typename X>
inline X sqr(const X& a) {
  return a * a;
}
bool rev(pair<long long int, long long int>& a,
         pair<long long int, long long int>& b) {
  return a.first < b.first || (a.first == b.first && a.second > b.second);
}
int dx[4] = {-1, +1, 0, 0};
int dy[4] = {0, 0, +1, -1};
int DX[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int DY[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
const int N = 505;
void test() {
  long long int k, n;
  cin >> n >> k;
  if (k == 1) {
    cout << n << endl;
    return;
  }
  long long int res = 1;
  while (res <= n) res *= 2;
  cout << res - 1 << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    test();
  }
  return 0;
}
