//適度にばらついて移動することを期待したDP(嘘解法)
#include <iostream>
#include <algorithm>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll = long long;
const long long INF = 1000000000000000000;
//#define endl "\n";
ll Q, N, K, A[200009][2];
const ll Kmax = 600;
const ll center = 300;

int main(){
  cin >> Q;
  for(int q = 1; q <= Q; q++){
    cin >> N >> K;
    for(int i = 0; i < N; i++) cin >> A[i][0];
    for(int i = 0; i < N; i++) cin >> A[i][1];
    ll ans = 0;

    vector<vector<ll>> dp(Kmax + 1, vector<ll>(2,-INF));
    vector<vector<ll>> nexdp(Kmax + 1, vector<ll>(2,-INF));
    ll K2 = (K + 1) / 2;
    dp[center][0] = 0;
    ll prehiku = 0;
      for(int i = 0; i < N; i++){
      ll hiku = K2 * (i + 1) / N;
      for(int j = 0; j <= Kmax; j++){
        if(prehiku == hiku){
          nexdp[j][0] = max(nexdp[j][0], dp[j][0] + A[i][0]);
          nexdp[j][1] = max(nexdp[j][1], dp[j][1] + A[i][1]);
          if(j + 1 <= Kmax) nexdp[j + 1][1] = max(nexdp[j + 1][1], dp[j][0] + A[i][1]);
          nexdp[j][0] = max(nexdp[j][0], dp[j][1] + A[i][0]);
        }else{
          //情報を1つ捨てる
          if(0 <= j - 1) nexdp[j - 1][0] = max(nexdp[j - 1][0], dp[j][0] + A[i][0]);
          if(0 <= j - 1) nexdp[j - 1][1] = max(nexdp[j - 1][1], dp[j][1] + A[i][1]);
          nexdp[j][1] = max(nexdp[j][1], dp[j][0] + A[i][1]);
          if(0 <= j - 1) nexdp[j - 1][0] = max(nexdp[j - 1][0], dp[j][1] + A[i][0]);
        }
      }
      prehiku = hiku;
      swap(dp, nexdp);
    }
    for(int j = max(0LL, center - K2); j < center; j++){
      for(int k = 0; k < 2; k++) ans = max(ans, dp[j][k]);
    }
    if(K2 * 2 == K) ans = max(ans, dp[center][0]);
    ans = max(ans, dp[center][1]);
    cout << ans << endl;
  }
  return 0;
}