You and your n - 1 friends have found an array of integers a_1, a_2, ..., a_n. You have decided to share it in the following way: All n of you stand in a line in a particular order. Each minute, the person at the front of the line chooses either the first or the last element of the array, removes it, and keeps it for himself. He then gets out of line, and the next person in line continues the process.

You are standing in the m-th position in the line. Before the process starts, you may choose up to k different people in the line, and persuade them to always take either the first or the last element in the array on their turn (for each person his own choice, not necessarily equal for all people), no matter what the elements themselves are. Once the process starts, you cannot persuade any more people, and you cannot change the choices for the people you already persuaded.

Suppose that you're doing your choices optimally. What is the greatest integer x such that, no matter what are the choices of the friends you didn't choose to control, the element you will take from the array will be greater than or equal to x?

Please note that the friends you don't control may do their choice arbitrarily, and they will not necessarily take the biggest element available.

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains three space-separated integers n, m and k (1 ≤ m ≤ n ≤ 3500, 0 ≤ k ≤ n - 1) — the number of elements in the array, your position in line and the number of people whose choices you can fix.

The second line of each test case contains n positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 3500.

Output

For each test case, print the largest integer x such that you can guarantee to obtain at least x.

Example

Input


4
6 4 2
2 9 2 3 8 5
4 4 1
2 13 60 4
4 1 3
1 2 2 1
2 2 0
1 2


Output


8
4
1
1

Note

In the first test case, an optimal strategy is to force the first person to take the last element and the second person to take the first element.

  * the first person will take the last element (5) because he or she was forced by you to take the last element. After this turn the remaining array will be [2, 9, 2, 3, 8]; 
  * the second person will take the first element (2) because he or she was forced by you to take the first element. After this turn the remaining array will be [9, 2, 3, 8]; 
  * if the third person will choose to take the first element (9), at your turn the remaining array will be [2, 3, 8] and you will take 8 (the last element); 
  * if the third person will choose to take the last element (8), at your turn the remaining array will be [9, 2, 3] and you will take 9 (the first element). 



Thus, this strategy guarantees to end up with at least 8. We can prove that there is no strategy that guarantees to end up with at least 9. Hence, the answer is 8.

In the second test case, an optimal strategy is to force the first person to take the first element. Then, in the worst case, both the second and the third person will take the first element: you will end up with 4.

// Problem : A. Mind Control
// Contest : Codeforces - Codeforces Round #616 (Div. 1)
// URL : https://codeforces.com/problemset/problem/1290/A
// Memory Limit : 256 MB
// Time Limit : 1000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
#define endl '\n'
#define maxl __LONG_LONG_MAX__
#define pb push_back
#define F first
#define S second
#define mp make_pair
#define rep(i, n) for(int i = 0; i < n; i++)
#define rev(i, n) for(int i = n-1; i >= 0; i--)
#define rep1(i, n) for(int i=1; i < n; i++)
#define rev1(i, n) for(int i=n ; i > 0; i--)
#define FOR(i, a, b) for(int i=a ; i < b; i++)
#define trace(x)                 cerr << #x << ": " << x << endl;
#define trace2(x, y)             cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
#define trace3(x, y, z)          cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
#define trace4(a, b, c, d)       cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
#define cout1(a) cout<<a<<endl
#define cout2(a,b) cout<<a<<' '<<b<<endl
#define cout3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define cout4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define coutvp(v,i) cout<<v[i].fi<<" "<<v[i].se<<endl
#define couta(a,n) for(ll i=0;i<n;i++){ cout<<a[i]<<" ";}
#define n_ones(x) __builtin_popcountll(x)
#define n_trailzero(x) __builtin_ctzll(x)
#define n_leadingzero(x) __builtin_clz(x)
#define sz(a) (ll)(a.size())
#define all(v) v.begin(),v.end()
#define FASTIO {ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);}
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) (a/gcd(a, b))*b
#define PI 3.14159265358979323846
#define see(args...)                       \
{                                            \
    cerr<<"LINE "<<__LINE__;                 \
    string _s = #args;                       \
    replace(_s.begin(), _s.end(), ',', ' '); \
    stringstream _ss(_s);                    \
    istream_iterator<string> _it(_ss);       \
    err(_it, args);                          \
    cerr<<endl;                              \
}
void err(istream_iterator<string> it){}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args)
{
  cerr<<' '<< *it << " = " << a ;
  err(++it, args...);
} 
template <typename T, typename U>inline void min_self(T &x, U y){if (y < x)x = y;}
template <typename T, typename U>inline void max_self(T &x, U y){if (x < y)x = y;}
template<class T1, class T2>ostream& operator<< (ostream &out, pair<T1, T2> pair) { return out << "(" << pair.first << ", " << pair.second << ")";}
template<class T> ostream& operator<<(ostream& out, vector<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }
template<class T> ostream& operator<<(ostream& out, set<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }
template<class L, class R> ostream& operator<<(ostream& out, map<L, R> vec) { out<<"("; for (auto& v: vec) out<<"["<<v.first<<", "<<v.second<<"]"; return out<<")"; }
template<class A,class B>istream&operator>>(istream&in,pair<A,B>&a){return in>>a.first>>a.second;}
template<class A>istream&operator>>(istream&in,vector<A>&a){for(A &i:a)in>>i;return in;}


typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef vector<ll> vll;
typedef vector<pair<ll, ll> > vpll;
typedef vector<vector<ll> > vvll;
typedef stack<ll> stk;
typedef queue<ll> que;

typedef tree<int,null_type,less<int>,rb_tree_tag,
tree_order_statistics_node_update> indexed_setll;
typedef tree<char,null_type,less<char>,rb_tree_tag,
tree_order_statistics_node_update> indexed_setc;

ll max(int a,ll b){ return max((ll)a,b);}
ll min(int a,ll b){ return min((ll)a,b);}
ll min(ll a,int b){ return min(a,(ll)b);}
ll max(ll a,int b){ return max(a,(ll)b);}

// cout<<fixed<<setprecision(20)<<pi<<endl; 

ll mod = 1e9 + 7;


void solve()
{
	ll n, m, k;
	cin>>n>>m>>k;
	vll A(n);
	rep(i, n)
		cin>>A[i];
	ll l = 0, r = n-1, cnt = 0, ans = 0;
	while(cnt < m)
	{
		cnt++;
		//see(cnt, l, r, A[l], A[r], k, ans);
		if(cnt == m)
			ans = max(A[l], A[r]);
		else if(k > 0)
		{
			k--;
			if(A[l] < A[r])
				l++;
			else 	
				r--;
		}
		else
		{
			if(A[l] > A[r])
				l++;
			else
				r--;
		}
	}
	cout1(ans);
	return;
}

int main()
{
	FASTIO;
	ll t;
	cin>>t;
	rep(i, t)
		solve();
	return 0;
}


