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.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static  ArrayList<Pair>adj[];
	static PrintWriter out=new PrintWriter(System.out);
	 public static int mod = 1000000007;

	static int notmemo[][][][];
	static int k;
	static int a[];
	static int b[];
	static int m;
	static char c[];
	
	static int trace[];
	static int h;
	static int x;
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		label:while(t-->0) {
			int n=sc.nextInt();
			int m=sc.nextInt();
			int k=sc.nextInt();
			int a[]=new int[n];
			for (int i = 0; i < a.length; i++) {
				a[i]=sc.nextInt();
			}
			int i=0;
			int j=n-1;
			int ans=0;
			int turn=0;
			while(i<n&&j>=0) {
				turn++;
				if(turn==m) {
					ans=Math.max(a[i], a[j]);
				}
				if(k>0) {
					k--;
					if(a[i]>a[j]) {
						j--;
						continue;
					}
					else {
						i++;
						continue;
					}
				}
				if(a[i]>a[j]) {
					i++;
					continue;
				}
				else {
					j--;
					continue;
				}
				
				
				
				
			}
			System.out.println(ans);
		}
		
		
	out.flush();
	}
	static class SegmentTree {		// 1-based DS, OOP
		
		int N; 			//the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;
		
		SegmentTree(int[] in)		
		{
			array = in; N = in.length - 1;
			sTree = new int[N<<1];		//no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N<<1];
			build(1,1,N);
		}
		
		void build(int node, int b, int e)	// O(n)
		{
			if(b == e)					
				sTree[node] = array[b];
			else						
			{
				int mid = b + e >> 1;
				build(node<<1,b,mid);
				build(node<<1|1,mid+1,e);
				sTree[node] = sTree[node<<1]+sTree[node<<1|1];
			}
		}
		
		
		void update_point(int index, int val)			// O(log n)
		{
			index += N - 1;				
			sTree[index] += val;			
			while(index>1)				
			{
				index >>= 1;
				sTree[index] = sTree[index<<1] + sTree[index<<1|1];		
			}
		}
		
		
		void update_range(int i, int j, int val)		// O(log n) 
		{
			update_range(1,1,N,i,j,val);
		}
		
		void update_range(int node, int b, int e, int i, int j, int val)
		{
			if(i > e || j < b)		
				return;
			if(b >= i && e <= j)		
			{
				sTree[node] += (e-b+1)*val;			
				lazy[node] += val;				
			}							
			else		
			{
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node<<1,b,mid,i,j,val);
				update_range(node<<1|1,mid+1,e,i,j,val);
				sTree[node] = sTree[node<<1] + sTree[node<<1|1];		
			}
			
		}
		void propagate(int node, int b, int mid, int e)		
		{
			lazy[node<<1] += lazy[node];
			lazy[node<<1|1] += lazy[node];
			sTree[node<<1] += (mid-b+1)*lazy[node];		
			sTree[node<<1|1] += (e-mid)*lazy[node];		
			lazy[node] = 0;
		}
		
		int query(int i, int j)
		{
			return query(1,1,N,i,j);
		}
		
		int query(int node, int b, int e, int i, int j)	// O(log n)
		{
			if(i>e || j <b)
				return 0;		
			if(b>= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			int q1 = query(node<<1,b,mid,i,j);
			int q2 = query(node<<1|1,mid+1,e,i,j);
			return q1 + q2;	
					
		}
		
		
	}
static long memo[][][];
	static long dp(int i,int not,int total) {
	
	if(not>=k||total>x) {
		
		return -INF;
	}
	 if(i==n&&total==x) {
		return 0;
   }
	else if(i==n) {
		return -INF;
	}

	if(memo[i][not][total]!=-1){
		return memo[i][not][total];
	}
	long take=-INF;
	 take=dp(i+1,0,total+1)+a[i];
	long leave=-INF;
	 leave=dp(i+1,not+1,total);
	return memo[i][not][total]=Math.max(leave, take);
	}
		static long nCr1(int n , int k)
	{
		if(n < k)
			return 0;
		if(k == 0 || k == n)
			return 1;
		if(comb[n][k] != -1)
			return comb[n][k];
		if(n - k < k)
			return comb[n][k] = nCr1(n, n - k);
		return comb[n][k] = ((nCr1(n - 1, k - 1)%mod) +( nCr1(n - 1, k)%mod))%mod;
	}
	
	
static class UnionFind {                                              
	int[] p, rank, setSize;
	int numSets;
	int max[];
	public UnionFind(int N) 
	{
		max=new int[N];
		p = new int[numSets = N];
		rank = new int[N];
		setSize = new int[N];
		for (int i = 0; i < N; i++) {  p[i] = i; setSize[i] = 1;
		max[i]=i;
		}
	}

	public int findSet(int i) { return p[i] == i ? i : (p[i] = findSet(p[i])); }

	public boolean isSameSet(int i, int j) { return findSet(i) == findSet(j); }

	public void unionSet(int i, int j) 
	{ 
		if (isSameSet(i, j)) 
			return;
		numSets--; 
		int x = findSet(i), y = findSet(j);
		if(rank[x] > rank[y]) { p[y] = x; setSize[x] += setSize[y];
		max[x]=Math.max(max[x], max[y]);
		
		}
		else
		{	p[x] = y; setSize[y] += setSize[x];
			if(rank[x] == rank[y]) rank[y]++; 
			max[y]=Math.max(max[x],max[y]);
			
		} 
	}

	private int getmax(int j) {
	return max[findSet(j)];
	}

	public int numDisjointSets() { return numSets; }

	public int sizeOfSet(int i) { return setSize[findSet(i)]; }
}









/**private static void trace(int i, int time) {
if(i==n) 
	return;


long ans=dp(i,time);
if(time+a[i].t<a[i].burn&&(ans==dp(i+1,time+a[i].t)+a[i].cost)) {
	
	trace(i+1, time+a[i].t);
    
	l1.add(a[i].idx);
    return;
}
    trace(i+1,time);

}**/



	static class incpair implements Comparable<incpair>
{
	int a; long b;
	int idx;
	incpair(int a, long dirg,int i) {this.a = a;	b = dirg; idx=i; }
	
	public int compareTo(incpair e){ return (int) (b - e.b);	}
}
	static class decpair implements Comparable<decpair>
{
	int a; long b;int idx;
	
	decpair(int a, long dirg,int i) {this.a = a;	b = dirg; idx=i;}
	
	public int compareTo(decpair e){ return (int) (e.b- b);	}
}

	
	
	
	
	
	


	static long allpowers[];
	static class Quad implements Comparable<Quad>{
		int u;
		int v;
		char state;
		int turns;
		public Quad(int i, int j, char c, int k) {
		u=i;
		v=j;
		state=c;
		turns=k;
		}
		public int compareTo(Quad e){ return (int) (turns - e.turns);	}
		
	}
	static long dirg[][];
	static Edge[] driver;
	
	static int n;
		static class Edge implements Comparable<Edge>
	{
		int node; long cost;
		
		Edge(int a, long dirg) { node = a;	cost = dirg; }
		
		public int compareTo(Edge e){ return (int) (cost - e.cost);	}
	}

	



	  static long manhatandistance(long x,long x2,long y,long y2) {
		  return  Math.abs(x-x2)+Math.abs(y-y2);
	  }
	  static long fib[];
	  static long fib(int n) {
		  if(n==1||n==0) {
			  return 1;
		  }
		  if(fib[n]!=-1) {
			  return fib[n];
		  }
		  else 
			  return  fib[n]=  ((fib(n-2)%mod+fib(n-1)%mod)%mod);
	  }
	  static class Pair implements Comparable<Pair>{
		  int v;
		  long cost;
		  Pair(int a, long b){
			  v=a;
			  cost=b;
		  }
		public int compareTo(Pair p) {
			return (int) (this.cost-p.cost);
		 
				 
		}
		  
		  
	
		
		
	}
	static long[][] comb;		
	
	
	
static  class Triple implements Comparable<Triple>{

int l;
int r;
long cost;
int idx;
		public Triple(int a,int b,long l1,int l2) {
			l=a;
			r=b;
			cost=l1;
			idx=l2;
		}
		public int compareTo(Triple x)
		{
			if(l!=x.l || idx==x.idx)
				return l-x.l;
			return -idx;
		}
		
		
	}
	
	static TreeSet<Long> primeFactors(long N)		// O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(p * p <= N)
		{
			while(N % p == 0) { factors.add((long) p); N /= p; }
			if(primes.size()>idx+1)
			p = primes.get(++idx);
			else
				break;
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}
	
	static boolean visited[];
	/**static int bfs(int s)
	{
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		int count=0;
		int maxcost=0;
		int dist[]=new int[n];
		dist[s]=0;
		while(!q.isEmpty())
		{
			
			int u = q.remove();
			if(dist[u]==k) {
				break;
			}
			for(Pair v: adj[u])
			{	
				maxcost=Math.max(maxcost, v.cost);
			
			
			
			if(!visited[v.v])	{
				
				visited[v.v]=true;
				q.add(v.v);
				dist[v.v]=dist[u]+1;
				maxcost=Math.max(maxcost, v.cost);
			}
			}	
			
		}
		return maxcost;
	}**/
	public static boolean FindAllElements(int n, int k) {
	 int sum = k;  
     int[] A = new int[k];  
      Arrays.fill(A, 0, k, 1);  
        
      for (int i = k - 1; i >= 0; --i) {  
      
           while (sum + A[i] <= n) {  
      
              sum += A[i];  
              A[i] *= 2;  
          }  
      }  
      if(sum==n) {
      	return true;
      }
      else
    	return false; 
	}
	

	
	static boolean vis2[][];
	
	
	static boolean f2=false;
	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r)	//C(p x r) = A(p x q) x (q x r) -- O(p x q x r)
	{
		long[][] C = new long[p][r];
		for(int i = 0; i < p; ++i) {
			for(int j = 0; j < r; ++j) {
				for(int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j]+(a2[i][k]%mod * b[k][j]%mod))%mod;
		 C[i][j]%=mod;
				}
		
			}
		}
		return C;
	}
		
	
	
		
		
		
		

		
	
	static ArrayList<Pair> a1[];
	static int memo1[];
	
	static boolean vis[];
	static TreeSet<Integer> set=new TreeSet<Integer>();
	
	static long modPow(long ways, long count, long mod)	// O(log e)
	{
		ways %= mod;
		long res = 1;
		while(count > 0)
		{
			if((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res%mod;
	}
	
	
	

	static long gcd(long ans,long b) {
		if(b==0) {
			return ans;
		}
		return gcd(b,ans%b);
	}
	static int[] isComposite;
	static int[] valid;
	
	static ArrayList<Integer> primes;
	static ArrayList<Integer> l;
	static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
		l=new ArrayList<>();
		valid=new int[N+1];
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				l.add(i);
				valid[i]=1;
				
				for (int j = i*2; j <=N; j +=i) {	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
						
						
						
				      
					}
			}
		
		for(int i=0 ; i<primes.size() ; i++) {
			for(int j:primes) {
				if(primes.get(i)*j>N) {
					break;
				}
				valid[primes.get(i)*j]=1;
			}
		}
	}
	
	public static long[] schuffle(long[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x=(int)(Math.random()*a2.length);
		long temp=a2[x];
		a2[x]=a2[i];
		a2[i]=temp;
		}
		return a2;
	}
		static int V;
	static long INF=(long) 1E16;
		
		static class Edge2 
		{
			int node;
		long cost;
			long next;
			Edge2(int a, int c,Long long1) { node = a;	cost = long1; next=c;}
			
		}
static class Scanner {
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}	
}