#include <bits/stdc++.h>
#include <chrono>
using namespace std;
using namespace chrono;
// "AJEET JAIN"----"JAI JINENDRA"
/* "णमो अरिहंताणं",
"णमो सिद्धाणं",
"णमो आयरियाणं",
"णमो उवज्झायाणं",
"णमो लोए सव्वसाहूणं",
"",
"एसो पंच नमोक्कारो, सव्व पावप्पणासणो",
"मंगलाणं च सव्वेसिं, पडमं हवै मंगलं", */
// Aliases to op
using ll = long long;
using ull = unsigned long long;
using ld = double;
using vll = vector<ll>;
// Constants
constexpr ll INF = 4e18;
constexpr ld EPS = 1e-9;
constexpr ll MOD = 1e9 + 7;
// Macros
#define F first
#define S second
#define all(x) begin(x), end(x)
#define allr(x) rbegin(x), rend(x)
#define py cout<<"YES\n";
#define pn cout<<"NO\n";
#define forn(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++)
// #define insert push_back
#define pb push_back
#define MP make_pair
#define endl '\n'
/*
remove substring or subarray ---> try to think about sliding w
*/
/*
Golden Rule
1) problem is easy
2) proofs is easy
3) implementation is easy
/*
ROUGH --
if u have a1 a2 a3 a4 a5
and b1 b2 b3 b4 b5
if x = 0 and y = 5 -> b1+b2+b3+b4+b5 = sum_of_b
if x = 1 and y = 4 -> sum_of_b + max(ai - bi)
...
*/
void AJNJ(){
int n, x , y;
cin >> n >> x >> y;
vector<ll> a(n) , b(n) , diff(n);
ll sum_b = 0;
for(int i = 0 ; i < n ; i++){
cin >> a[i];
}
for(int i = 0 ; i < n ; i++){
cin >> b[i];
sum_b += b[i];
}
for(int i = 0 ; i < n ; i++){
diff.push_back(a[i] - b[i]);
}
sort(allr(diff));
ll ans = INT_MIN;
ll sum = 0;
forn(i , n){
if(i + 1 <= x){
sum += diff[i];
ans = max(ans , sum_b + sum );
}
else{
break;
}
}
cout << ans << endl;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int T = 1;
cin>>T;
auto start1 = high_resolution_clock::now();
while(T--){
AJNJ();
}
auto stop1 = high_resolution_clock::now();
auto duration = duration_cast<microseconds>(stop1 - start1);
cerr << "Time: " << duration . count() / 1000 << " ms" << endl;
return 0;
}
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