#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
const int N=2e6 + 9;
ll n,m,k;
ll a[N],c[N];
ll dp[N];
bool check(ll x){
	ll res = 0;
	for(int i = 1;i <= (n + 1) / 2 ;i++){
		if(a[i] <= x) continue;
		if(x == 0){
			res += dp[a[i] + 1];
			continue;
		}
		int K = (a[i]+x-1)/x;
		
		res += dp[K];
	}
	if(res <= k) return 0;
	return 1;
}
void solve(){
	cin>>n>>m>>k;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	for(int i=1;i<=m;i++){
		cin>>c[i];
		dp[i] = c[i];
		dp[i + m] = 1e16;
	}
	dp[1] = 0;
	sort(a + 1,a + n + 1);
	for(int i = m;i >= 1;i --) dp[i] = min(dp[i + 1] , dp[i]);
 	for(int i = 2;i <= 2 * m;i++){
		for(int j = 1;j * i <= 2 * m;j++)
		{
			dp[j * i] = min(dp[j * i], dp[i] + dp[j]);
		}
	}
	for(int i = 2 * m - 1;i >= 1;i --){
		dp[i] = min(dp[i + 1], dp[i]);
	}
//	for(int i=1;i<=m;i++){
//		cout<<"dp["<<i<<"]="<<dp[i]<<endl;
//	}
	// 除 x 所需要的最小代价
	// 除至少为 x 的最小代价
	ll l = 0 ,r = 2 * m + 1;
	while(l<=r){
		ll mid=(l + r)>>1;
//		cout <<l <<" " <<r <<" "<<mid<<endl;
		if(check(mid))l = mid + 1;
		else r = mid - 1;
	}
	cout << r + 1 <<endl;
}
int main(){
	int _=1;
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>_;
	while(_--)
	solve();
	return 0;
}