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.
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Problem651D {
	public static void main(String[] args) throws Exception {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter writer = new PrintWriter(System.out);

		StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
		int n = Integer.parseInt(tokenizer.nextToken());
		int k = Integer.parseInt(tokenizer.nextToken());
		long[] a = new long[n];
		StringTokenizer aT = new StringTokenizer(reader.readLine());
		for (int i = 0; i < n; i++) {
			a[i] = Long.parseLong(aT.nextToken());
		}
		writer.println(binarySearch(a, k));

		reader.close();
		writer.close();
	}
	public static long binarySearch(long[] array, int k) {
		long low = 0;
		long high = 1000000005;
		while (low < high) {
			if (high - low <= 1) {
				if (possible(array, low, k)) {
					return low;
				} else {
					return high;
				}
			}
			long mid = (high - low) / 2 + low;
			if (possible(array, mid, k)) {
				high = mid;
			} else {
				low = mid + 1;
			}
		}
		return low;
	}
	public static boolean possible(long[] array, long value, int k) {
		return possible1(array, value, k) || possible2(array, value, k);
	}
	public static boolean possible1(long[] array, long value, int k) {
		int lastIndex = -99;
		int count = 0;
		int firstIndex = -1;
		for (int i = 0; i < array.length; i++) {
			if (array[i] <= value) {
				if (lastIndex == i - 1) {
					continue;
				} else {
					if (firstIndex == -1) {
						firstIndex = i;
					}
					lastIndex = i;
					count++;
				}
			}
		}
		int sum = count + (count - 1);
		if (firstIndex > 0) {
			sum++;
		}
		if (lastIndex < array.length - 1) {
			sum++;
		}
		return sum >= k;
	}
	public static boolean possible2(long[] array, long value, int k) {
		int lastIndex = -99;
		int count = 0;
		int firstIndex = -1;
		// skip first element
		for (int i = 1; i < array.length; i++) {
			if (array[i] <= value) {
				if (lastIndex == i - 1) {
					continue;
				} else {
					if (firstIndex == -1) {
						firstIndex = i;
					}
					lastIndex = i;
					count++;
				}
			}
		}
		int sum = count + (count - 1);
		if (firstIndex > 0) {
			sum++;
		}
		if (lastIndex < array.length - 1) {
			sum++;
		}
		return sum >= k;
	}
}
