Codeforces C. Less or Equal

C. Less or Equal

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequence of integers of length $n$ and integer number $k$. You should print any integer number $x$ in the range of $[1;{10}^9]$ (i.e. $1\le x\le {10}^9$) such that exactly $k$ elements of given sequence are less than or equal to $x$.

Note that the sequence can contain equal elements.

If there is no such $x$, print "-1" (without quotes).

Input

The first line of the input contains integer numbers $n$ and $k$ ($1\le n\le 2\cdot {10}^5$, $0\le k\le n$). The second line of the input contains $n$ integer numbers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le {10}^9$) — the sequence itself.

Output

Print any integer number $x$ from range $[1;{10}^9]$ such that exactly $k$ elements of given sequence is less or equal to $x$.

If there is no such $x$, print "-1" (without quotes).

Examples

input

Copy

7 4
3 7 5 1 10 3 20

output

Copy

6

input

Copy

7 2
3 7 5 1 10 3 20

output

Copy

-1

Note

In the first example $5$ is also a valid answer because the elements with indices $[1,3,4,6]$ is less than or equal to $5$ and obviously less than or equal to $6$.

In the second example you cannot choose any number that only $2$ elements of the given sequence will be less than or equal to this number because $3$ elements of the given sequence will be also less than or equal to this number.

题目大意：

给你一个n个数字组成的序列，问是否存在一个整数x，使得该序列刚好有k个数小于或等于x，存在即输出该x，否则输出-1。

解题思路：

先对序列排序，x先定为排序后的第k个数，然后对序列遍历一次，如果小于等于x的的个数刚好等于k个，说明x为所求，否则不存在这样的x。另外还要考虑k=0的情况，即要找到这样一个x使该序列没有元素小于它，如果该序列最小元素为1，那1肯定是小于等于任何整数x，所以不存在。否则x为最小元素减1。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.StringTokenizer;

public class LessorEqual {
	static class InputReader {
	    public BufferedReader reader;
	    public StringTokenizer tokenizer;
	
	    public InputReader(InputStream stream) {
	        reader = new BufferedReader(new InputStreamReader(stream), 32768);
	        tokenizer = null;
	    }
	
	    public String next() {
	        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }
	
	    public int nextInt() {
	        return Integer.parseInt(next());
	    }
	
	}
	public static void main(String[] args) {
		InputReader sc = new InputReader(System.in);
		int n,k;
		n = sc.nextInt();
		k = sc.nextInt();
		Integer[] a = new Integer[n];
		for(int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		Arrays.sort(a);
		int ans = 0;
		if(k == 0)ans = a[0] - 1;
		else ans = a[k-1];
		int cnt = 0;
		for(int i = 0; i < n; i++) {
			if (a[i] <= ans) cnt++;
		}
		if(cnt != k || !(ans >= 1 && ans <= 1000*1000*1000))ans = -1;
		System.out.print(ans);
	}
}