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.
from collections import defaultdict
(n, k)=[int(x) for x in input().split()]

# res =-1
# dp = [1] * (n+1)
# for i in range(n,0,-1):
#     if dp[i] >= k:
#         res = i
#         break
#     imod2 = i % 2
#     if imod2 == 0:
#         dp[i//2] += dp[i]
#     else:            
#         dp[i-1] += dp[i]
#         
# print(dp)
# print(res)

# res =-1
# dp = defaultdict(lambda:1)
# for i in range(n,0,-1):
#      val = dp[i]
#      if val >= k:
#          res = i
#          break    
#      imod2 = i % 2
#      idx =-1
#      if imod2 == 0:         
#          idx = i//2
#      else:
#          idx = i-1
#      dp[idx] += val
#      del dp[i]
# print(dp)
# print(res)


#import snoop
#@snoop
def f(x):
    res = 1
    if x % 2 == 1:
        adj = 1
    else:
        adj = 2    
    res = adj    
    i = 2
    while i*x <= n:
        if i*x + adj * i - 1 <=n:
            res+= i * adj
            i*=2
        else:
            res+= (n- i*x + 1)
            break
    return res            

#import snoop
#@snoop
def find():
    lo=1
    hi=n//2
    re=1
    while lo<=hi:
        m=(lo+hi)//2;
        if f(2*m)>=k:
            re=2*m
            lo=m+1    
        else:
            hi=m-1
    lo=1
    hi=n//2 + n % 2
    ro=1
    while lo<=hi:
        m=(lo+hi)//2;
        if f(2*m - 1)>=k:
            ro=2*m -1
            lo=m+1    
        else:
            hi=m-1
    return max(re,ro)            
print(find())
