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;
long long int INF = 1e18;
const long long fact_table = 5000000;
double Pi = 3.1415926535897932384626;
vector<long long> G[550010];
vector<pair<long long, double> > tree[500010];
priority_queue<long long> pql;
priority_queue<pair<long long, long long> > pqp;
priority_queue<long long, vector<long long>, greater<long long> > pqls;
priority_queue<pair<long long, long long>, vector<pair<long long, long long> >,
               greater<pair<long long, long long> > >
    pqps;
long long dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
long long dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
char dir[] = "DRUL";
long long mod = 1000000007;
long long rui(long long number1, long long number2) {
  if (number2 == 0) {
    return 1;
  } else {
    long long number3 = rui(number1, number2 / 2);
    number3 *= number3;
    number3 %= mod;
    if (number2 % 2 == 1) {
      number3 *= number1;
      number3 %= mod;
    }
    return number3;
  }
}
long long gcd(long long number1, long long number2) {
  if (number1 > number2) {
    swap(number1, number2);
  }
  if (number1 == 0 || number1 == number2) {
    return number2;
  } else {
    return gcd(number2 % number1, number1);
  }
}
void YES(bool condition) {
  if (condition) {
    cout << "YES"
         << "\n";
    ;
  } else {
    cout << "NO"
         << "\n";
    ;
  }
  return;
}
void Yes(bool condition) {
  if (condition) {
    cout << "Yes"
         << "\n";
    ;
  } else {
    cout << "No"
         << "\n";
    ;
  }
  return;
}
long long n, m, num, sum, a, b, c, d, e, f, g, h, w, i, j, q, r, l;
long long k, ans;
long long x[500005], y[500005], z[500005];
long long reside[500005], memo[500005];
bool used[500005], got[400004];
char s[500005];
long long count(long long num) {
  long long cnt = 0;
  if (num == 1) {
    return n;
  }
  if (num == 2) {
    return n - 1;
  }
  if (num % 2 == 1) {
    cnt++;
    num = 2 * num;
  }
  long long low = num, high = num + 1;
  while (true) {
    if (low > n) break;
    cnt += min(high, n) - low + 1;
    low *= 2;
    high = high * 2 + 1;
  }
  return cnt;
}
long long solve_even() {
  long long ok = 0, ng = INF / 2;
  while (ng - ok > 1) {
    long long mid = (ok + ng) / 2;
    long long cnt = count(mid * 2);
    if (cnt >= k) {
      ok = mid;
    } else {
      ng = mid;
    }
  }
  return ok * 2;
}
long long solve_odd() {
  long long ok = 0, ng = INF / 2;
  while (ng - ok > 1) {
    long long mid = (ok + ng) / 2;
    long long cnt = count(mid * 2 + 1);
    if (cnt >= k) {
      ok = mid;
    } else {
      ng = mid;
    }
  }
  return ok * 2 + 1;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> k;
  if (k == 1) {
    cout << n << "\n";
    ;
    return 0;
  }
  long long even, odd;
  even = solve_even();
  odd = solve_odd();
  ans = max(even, odd);
  cout << ans << "\n";
  ;
  return 0;
}
