CodeForces 977 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 to 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.

给n个数，找出一个数，使有k个数小于等于该数。

思路一想就能出来，注意k=0的情况。

AC代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
int a[200010];
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        int t;
        if(!k)
            cout<<(t=a[0]-1>=1?a[0]-1:-1)<<endl;
        else if(a[k-1]!=a[k]) cout<<a[k-1]<<endl;
        else cout<<-1<<endl;
    }
    return 0;
}