Less or Equal ——NEUQ-ACM Codeforces Round #479 (Div. 3)

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我们先回顾下题目

C - Less or Equal

 

You are given a sequence of integers of length nn and integer number kk. You should print any integer number xx in the range of [1;109][1;109] (i.e. 1≤x≤1091≤x≤109) such that exactly kk elements of given sequence are less than or equal to xx.

Note that the sequence can contain equal elements.

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

Input

The first line of the input contains integer numbers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, 0≤k≤n0≤k≤n). The second line of the input contains nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the sequence itself.

Output

Print any integer number xx from range [1;109][1;109] such that exactly kk elements of given sequence is less or equal than xx.

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

Examples

Input

7 4
3 7 5 1 10 3 20

Output

6

Input

7 2
3 7 5 1 10 3 20

Output

-1

Note

In the first example 55 is also a valid answer because the elements with indices [1,3,4,6][1,3,4,6] is less than or equal to 55 and obviously less than or equal to 66.

In the second example you cannot choose any number that only 22 elements of the given sequence will be less than or equal to this number because 3

3 elements of the given sequence will be also less than or equal to this number.

这道题的精髓就是注意数据的边界问题。

要注意的坑：当k=0时和k=n时，需要额外讨论！最后就是判断普通情况下不能取到x的值的情况。

代码如下：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    int num[n];
    for(int i=0;i<n;++i)
    {
        scanf("%d",&num[i]);
    }
    sort(num,num+n);
    if(k==0)
    {
        if(num[0]>=2) printf("%d\n",num[0]-1);
        else printf("-1\n");
        return 0;
    }
    if(k==n)
    {
        printf("%d\n",num[n-1]);
        return 0;
    }

    if(num[k]==num[k-1])
        printf("-1\n");
    else printf("%d\n",num[k]-1);
    return 0;
}