不要手贱写树套树！！！

本题草场可以看成是线段, 我用线段树维护这些线段, 每个节点存一个 set 维护覆盖这个节点的线段.

每次查询全局最大值, 然后把覆盖这个单点的所有线段在树上抹去.

我这样的算法会在某些情况出问题, 如果有三条线段表示开区间:

$(1, 2)$, $(1, 7)$, $(4, 10)$, $(9, 10)$.

如果我一开始选择了线段 $(1, 7)$, $(4, 10)$, 之后需要再选两次才能选完这些线段. 总共需要选 $3$ 次.

但是这 $4$ 条线段只需要两次就可以全部选完, 所以这个做法是错误的.

pair<unsigned, unsigned> Grass[200005];
unsigned Rg[200005][2], Cow[200005], b[400005];
unsigned Stack[200005], STop(0);
unsigned long long Ans(0);
unsigned m, n, My;
unsigned A, B, C, D, t;
unsigned Cnt(0), Tmp(0);
struct Node {
  set<unsigned> List;
  Node* LS, *RS;
  unsigned long long Mx, Tag;
  unsigned MxP;
}N[800005], *CntN(N);
inline void Build(Node*x, unsigned L, unsigned R) {
  x->MxP = L;
  if(L == R) return;
  unsigned Mid((L + R) >> 1);
  Build(x->LS = ++CntN, L, Mid);
  Build(x->RS = ++CntN, Mid + 1, R);
}
inline void Chg(Node* x, unsigned L, unsigned R) {
//  printf("Chg %u [%u, %u] for [%u, %u]\n", x - N, L, R, A, B);
  if((A <= L) && (R <= B)) {
    x->List.insert(C), x->Tag += Grass[C].second, x->Mx += Grass[C].second;
    return;
  }
  unsigned Mid((L + R) >> 1);
  if(A <= Mid) Chg(x->LS, L, Mid);
  if(B > Mid) Chg(x->RS, Mid + 1, R);
  if(x->LS->Mx >= x->RS->Mx) x->Mx = x->Tag + x->LS->Mx, x->MxP = x->LS->MxP;
  else x->Mx = x->Tag + x->RS->Mx, x->MxP = x->RS->MxP;
}
inline void Access(Node* x, unsigned L, unsigned R) {
//  printf("Access %u (%u) [%u, %u] LS %u RS %u\n", x - N, A, L, R, x->LS - N, x->RS - N);
  for (auto i:x->List) Stack[++STop] = i;
  if(L == R) return;
  unsigned Mid((L + R) >> 1);
  if(A <= Mid) Access(x->LS, L, Mid);
  else Access(x->RS, Mid + 1, R);
}
inline void Ers(Node* x, unsigned L, unsigned R) {
//  printf("Ers %u [%u, %u] for [%u, %u]\n", x - N, L, R, A, B);
  if((A <= L) && (R <= B)) {
//    printf("Erase %u\n", C);
    x->List.erase(C);
//    for (auto i:x->List) printf("%u ", i); putchar(0x0A);
    x->Tag -= Grass[C].second, x->Mx -= Grass[C].second;
    return;
  }
  unsigned Mid((L + R) >> 1);
  if(A <= Mid) Ers(x->LS, L, Mid);
  if(B > Mid) Ers(x->RS, Mid + 1, R);
  if(x->LS->Mx >= x->RS->Mx) x->Mx = x->Tag + x->LS->Mx, x->MxP = x->LS->MxP;
  else x->Mx = x->Tag + x->RS->Mx, x->MxP = x->RS->MxP;
}
//  inline void Clr() {}
signed main() {
  freopen("9.in", "r", stdin);
//  freopen(".out", "w", stdout);
//  t = RD();
//  for (unsigned T(1); T <= t; ++T){
//  Clr();
  n = RD(), m = RD(), My = RD();
  for (unsigned i(1); i <= n; ++i) Grass[i].first = RD() + 1, Grass[i].second = RD();
  for (unsigned i(1); i <= m; ++i) Cow[i] = RD() + 1;
  sort(Grass + 1, Grass + n + 1), sort(Cow + 1, Cow + m + 1);
  Cow[0] = 0, Cow[++m] = 1000000002;
  for (unsigned i(0), j(1); j <= n; ++j) {
    while(Cow[i + 1] < Grass[j].first) ++i;
    unsigned Dis;
    if((i ^ (m - 1)) && ((!i) || (Cow[i + 1] - Grass[j].first < Grass[j].first - Cow[i]))) {
      Rg[j][1] = Cow[i + 1]; 
      Dis = Cow[i + 1] - Grass[j].first;
      Rg[j][0] = (Dis <= Grass[j].first) ? (Grass[j].first - Dis + 1) : 1;
    }
    else {
      Rg[j][0] = Cow[i] + 1;
      Dis = Grass[j].first - Cow[i];
      Rg[j][1] = Grass[j].first + Dis;
    }
  }
  for (unsigned i(1); i <= n; ++i) b[++Cnt] = Rg[i][0], b[++Cnt] = Rg[i][1];
  sort(b + 1, b + Cnt + 1), Cnt = unique(b + 1, b + Cnt + 1) - b;
  printf("Cnt %u\n", Cnt);
  Build(N, 1, Cnt - 1);
  for (unsigned i(1); i <= n; ++i) {
    A = Rg[i][0] = lower_bound(b + 1, b + Cnt, Rg[i][0]) - b;
    B = Rg[i][1] = lower_bound(b + 1, b + Cnt, Rg[i][1]) - b;
    C = i, Chg(N, 1, Cnt - 1);
  }
//  for (auto i:Test) printf("%u ", i); putchar(0x0A);
  while (My && N->Mx) {
    Ans += N->Mx, ++D;
    A = N->MxP, Access(N, 1, Cnt - 1);
//    sort(Stack + 1, Stack + STop + 1), STop = unique(Stack + 1, Stack + STop + 1) - Stack - 1;
//    printf("This Time %llu (%u) Size %u\n", N->Mx, N->MxP, STop);
//    for (unsigned i(1); i <= STop; ++i) printf("%u ", Stack[i]); putchar(0x0A);
//    system("pause");
    for (unsigned i(1); i <= STop; ++i)
      C = Stack[i], A = Rg[C][0], B = Rg[C][1], Ers(N, 1, Cnt - 1);
    --My, STop = 0;
  }
  printf("%llu %u\n", Ans, D);
  return Wild_Donkey;
}