human
Write me a Python code to find the minimum cost of a subsequence of size k given the following:
Ashish has an array a of size n.

A subsequence of a is defined as a sequence that can be obtained from a by deleting some elements (possibly none), without changing the order of the remaining elements.

Consider a subsequence s of a. He defines the cost of s as the minimum between: 

  * The maximum among all elements at odd indices of s. 
  * The maximum among all elements at even indices of s. 


Note that the index of an element is its index in s, rather than its index in a. The positions are numbered from 1. So, the cost of s is equal to min(max(s_1, s_3, s_5, …), max(s_2, s_4, s_6, …)).

For example, the cost of \{7, 5, 6\} is min( max(7, 6), max(5) ) = min(7, 5) = 5.

Help him find the minimum cost of a subsequence of size k.

Input

The first line contains two integers n and k (2 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the size of the array a and the size of the subsequence.

The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array a.

Output

Output a single integer — the minimum cost of a subsequence of size k.

Examples

Input

4 2
1 2 3 4

Output

1

Input

4 3
1 2 3 4

Output

2

Input

5 3
5 3 4 2 6

Output

2

Input

6 4
5 3 50 2 4 5

Output

3

Note

In the first test, consider the subsequence s = \{1, 3\}. Here the cost is equal to min(max(1), max(3)) = 1.

In the second test, consider the subsequence s = \{1, 2, 4\}. Here the cost is equal to min(max(1, 4), max(2)) = 2.

In the fourth test, consider the subsequence s = \{3, 50, 2, 4\}. Here the cost is equal to min(max(3, 2), max(50, 4)) = 3.
N,k=m
assistant
#Here is Python code to find the minimum cost of a subsequence of size k.
ap(int,input().split())
a=list(map(int,input().split()))

if k%2==0:
    def cond(n):
        res=0
        L=0
        for i in range(N):
            if a[i]>n:
                res+=(L+1)//2
                L=0
            else:
                L+=1
        res+=L//2
        if res>=k//2:
            return True

        res=0
        L=0
        for i in range(N-1,-1,-1):
            if a[i]>n:
                res+=(L+1)//2
                L=0
            else:
                L+=1
        res+=L//2
        if res>=k//2:
            return True

        return False

    start=0
    end=10**9
    while end-start>1:
        test=(end+start)//2
        if cond(test):
            end=test
        else:
            start=test
    print(end)
else:
    def cond(n):
        res=0
        L=0
        for i in range(N):
            if a[i]>n:
                res+=(L+1)//2
                L=0
            else:
                L+=1
        res+=(L+1)//2
        if res>=k//2+1:
            return True

        res=0
        L=0
        for i in range(N-2,-1,-1):
            if a[i]>n:
                res+=(L+1)//2
                L=0
            else:
                L+=1
        res+=L//2
        if res>=k//2:
            return True

        return False
    start=0
    end=10**9
    while end-start>1:
        test=(end+start)//2
        if cond(test):
            end=test
        else:
            start=test
    print(end)