#include <iostream>
#include <algorithm>
#include <string>
#include <assert.h>
#include <map>
using namespace std;
 
 
 
long long readInt(long long l,long long r,char endd){
	long long x=0;
	int cnt=0;
	int fi=-1;
	bool is_neg=false;
	while(true){
		char g=getchar();
		if(g=='-'){
			assert(fi==-1);
			is_neg=true;
			continue;
		}
		if('0'<=g && g<='9'){
			x*=10;
			x+=g-'0';
			if(cnt==0){
				fi=g-'0';
			}
			cnt++;
			assert(fi!=0 || cnt==1);
			assert(fi!=0 || is_neg==false);
			
			assert(!(cnt>19 || ( cnt==19 && fi>1) ));
		} else if(g==endd){
			assert(cnt>0);
			if(is_neg){
				x= -x;
			}
			assert(l<=x && x<=r);
			return x;
		} else {
			assert(false);
		}
	}
}
string readString(int l,int r,char endd){
	string ret="";
	int cnt=0;
	while(true){
		char g=getchar();
		assert(g!=-1);
		if(g==endd){
			break;
		}
		cnt++;
		ret+=g;
	}
	assert(l<=cnt && cnt<=r);
	return ret;
}
long long readIntSp(long long l,long long r){
	return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
	return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
	return readString(l,r,'\n');
}
string readStringSp(int l,int r){
	return readString(l,r,' ');
}


int T;
int n,k;
int sm_n=0;
map<int,int> mp;
int h;
pair<int,int> arr[100100];
int sm[100100];
int m;
map<int,int>::iterator ii;
int main(){
	T=readIntLn(1,100);
	while(T--){
		mp.clear();
		n=readIntSp(1,100000);
		sm_n += n;
		assert(sm_n<=100000);
		k=readIntLn(0,n);
		k=n-k; // instead of selecting K sticks, we remove n-k sticks
		for(int i=0;i<n-1;i++){
			h=readIntSp(1,1000000000);
			mp[h]++; // keep frequencies of sticks in a map
		}
		h=readIntLn(1,1000000000);
		mp[h]++;
		m=1;
		for(ii=mp.begin();ii!=mp.end();ii++){
			if(ii->second > 1){
				arr[m++] = *ii; 
			}
		}
		m--;
		int req=0;
		int mxx=0;
		for(int i=1;i<=m;i++){
			sm[i] = sm[i-1] + arr[i].second; // prefix sum of frequencies, we gonna need that
			if(arr[i].second > 1){
				req += arr[i].second -1;
				mxx = max ( arr[i].second -1,mxx);
			}
		}
		if (req -min(mxx,2)<= k){ // condition to tell if we can prevent having any rectangle
			cout<<-1<<endl;
			continue;
		}
		long long ans=1ll<<60;
		for(int i=m;i>=1;i--){   // go through sticks from tallest to smallest
			if(arr[i].second > 3){ // if we leave more than 3 sticks of currently tallest stick then answer is immediately  arr[i].first * 1ll *arr[i].first
				if(arr[i].second - 3 > k){
					ans = min(ans, arr[i].first * 1ll *arr[i].first);
					break;
				} 
				k -= arr[i].second  - 3;
				arr[i].second= 3;
			}
			if(arr[i].second  == 3){ // if there are  3 sticks we don't know whether we should remove 2 sticks or not, sometimes first option is optimal sometimes second
				int l=1,r=i; // so if we decided to leave 3 sticks then Chef's brother clear will take 2 of them for first dimension of rectangle
				while(r-l>1){ // now we have to minimize the second dimension
					int mid=(r+l)/2;
					if(sm[i-1] - sm[mid-1] - (i-mid) > k){
						l=mid;
					} else {
						r=mid;
					}
				}
				ans = min(ans,arr[i].first * 1ll * arr[l].first);
			}
			if(arr[i].second > 1){ // if there are more than 1 stick we have to take them
				if(arr[i].second - 1 > k){ // if we can't take them then we minimize the second dimension
					for(int j=i-1;j>=1;j--){
						if(arr[j].second - 1 > k){
							ans = min (ans, arr[j].first * 1ll * arr[i].first);
							break;
						}
						k -= arr[j].second - 1;
					}
					break;
				} 
				k -= arr[i].second  - 1; // if we can take them we try this option
				arr[i].second = 1;
			}
		}
		cout<<ans<<endl;
	}
	assert(getchar()==-1);
}