#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main() {
  ll N, M, X;
  cin >> N >> M >> X;
  ll G, B;
  cin >> G;
  vector<ll> old(N+1, 0);//変換前のマス。0なら授業、1なら得単、-1なら落単
  vector<ll> g(G+1);
  for(int i = 1; i <= G; i++) {
    cin >> g.at(i);
    old.at(g.at(i)) = 1;
  }
  cin >> B;
  vector<ll> b(B+1);
  for(int i = 1; i <= B; i++) {
    cin >> b.at(i);
    old.at(b.at(i)) = -1;
  }

  vector<ll> count(N+1, 0);
  for(int i = 1; i <= N; i++) {
    if(old.at(i) == -1) {
      ll l = max(0LL, i-X);
      ll r = min(N, i+X);
      for(int j = l; j <= r; j++) {
        count.at(j)++;
      }
    }
  }

  vector<ll> nw(1, 0);//変換後のマスの情報
  for(int i = 1; i <= N; i++) {
    if(count.at(i) != 0) {
      nw.push_back(old.at(i));
    }
    else {
      ll sum = 0;
      sum += old.at(i);
      while(i + 1 <= N && count.at(i+1) == 0) {
        i++;
        sum += old.at(i);
      }
      nw.push_back(sum);
    }
  }

  ll N1 = nw.size() - 1;//変換後のマス目数
  vector<vector<ll>> dp(N1+1, vector<ll>(M+1, -100100100LL));
  dp.at(0).at(M) = 0;

  for(int i = 1; i <= N1; i++) {
    for(int j = 0; j <= M; j++) {
       if(j != M && i >= X) {
        dp.at(i).at(j) = max(dp.at(i).at(j), dp.at(i-X).at(j+1) + nw.at(i));
       }
       dp.at(i).at(j) = max(dp.at(i).at(j), dp.at(i-1).at(j) + nw.at(i));
    }
  }

  ll ans = -100100100LL;//単位の数はマイナスを取りうることに注意してください。
  for (int j = 0; j <= M; j++) {
    ans = max(ans, dp.at(N1).at(j));
  }

  cout << ans << endl;

}