RT, WA on 第15个点7000多行，不知道哪里有问题。。。

#include <bits/stdc++.h>
using namespace std;
#define N 600010
#define ll long long 
#define int long long
const int INF = 2147483647; 

template <class T>
inline void read(T& a){
  T x = 0, s = 1;
  char c = getchar(); 
  while(!isdigit(c)){
    if(c == '-') s = -1;
    c = getchar(); 
  }
  while(isdigit(c)){
    x = x * 10 + (c ^ '0');
    c = getchar(); 
  }
  a = x * s;
  return ; 
}

int n; 
int root[N]; 

struct Treap{
  struct node{
    int siz, val, pri; 
    int lson, rson; 
  } t[N * 50]; 

  int tot; 

  Treap(int tot = 0){
    this->tot = tot; 
    return ; 
  }

  int build(int key = 0){
    tot++; 
    t[tot].val = key; 
    t[tot].pri = rand() * rand() % (ll)1e9; 
    t[tot].lson = t[tot].rson = 0; 
    t[tot].siz = 1;
    return tot; 
  }

  inline void pushup(int now){
    t[now].siz = t[t[now].lson].siz + t[t[now].rson].siz + 1; 
    return ; 
  }

  void split(int now, int key, int &x, int &y){
    if(!now){
      x = y = 0;
      return ; 
    }
    if(key >= t[now].val){
      x = build();
      t[x] = t[now]; 
      split(t[x].rson, key, t[x].rson, y); 
      pushup(x); 
    }
    else{
      y = build(); 
      t[y] = t[now]; 
      split(t[y].lson, key, x, t[y].lson); 
      pushup(y); 
    }
    return ; 
  }

  int merge(int x, int y){
    if(!x || !y) return x + y; 
    if(t[x].pri > t[y].pri){
      int now = build();   // merge 是否新建节点取决于版本间的 merge 与 split 操作是否成对出现。
      t[now] = t[x];      // 如果成对出现，此处新不新建均可。
      t[now].rson = merge(t[now].rson, y); 
      pushup(now);
      return now; 
    }
    else{
      int now = build(); 
      t[now] = t[y]; 
      t[now].lson = merge(x, t[now].lson);
      pushup(now);
      return now; 
    }
  }

  void insert(int key, int k){
    int x, y; 
    split(root[k], key - 1, x, y); 
    root[k] = merge(x, merge(build(key), y)); 
    return ; 
  }

  void del(int key, int k){
    int x, y, z; 
    split(root[k], key - 1, x, z); 
    split(z, key, y, z); 
    if(y){
      y = merge(t[y].lson, t[y].rson); 
    }
    root[k] = merge(merge(x, y), z); 
    return ; 
  }

  int find_kth(int now, int k){
    if(!now) return -INF; 
    if(t[t[now].lson].siz + 1 == k) return t[now].val;   // 这句话一定放最前面，先判断。否则会一路下去
    else if(t[t[now].lson].siz >= k) return find_kth(t[now].lson, k); 
    else return find_kth(t[now].rson, k - t[t[now].lson].siz - 1); 
  }

  int find_rank(int key, int k){
    int x, y, ans; 
    split(root[k], key - 1, x, y); 
    ans = t[x].siz + 1; 
    root[k] = merge(x, y); 
    return ans; 
  }

  int get_pre(int key, int k){
    int x, y, ans, now; 
    split(root[k], key - 1, x, y); 
    now = x; 
    while(t[now].rson) now = t[now].rson; 
    ans = x ? t[now].val : -INF; 
    root[k] = merge(x, y); 
    return ans; 
  }

  int get_next(int key, int k){
    int x, y, ans, now; 
    split(root[k], key + 1, x, y); 
    now = y; 
    while(t[now].lson) now = t[now].lson; 
    ans = y ? t[now].val : INF; 
    root[k] = merge(x, y); 
    return ans; 
  }

  void dfs(int now){
    if(!now) return ; 
    dfs(t[now].lson);
    printf("%d ", t[now].val); 
    dfs(t[now].rson);
    return ; 
  }

} tree; 

signed main(){
  srand(time(0)); 
  freopen("hh.txt", "r", stdin); 
  freopen("out.txt", "w", stdout); 
  read(n);
  for(int i = 1; i <= n; i++){
    int v, opt, x; 
    read(v), read(opt), read(x); 
    root[i] = root[v]; 
    switch (opt){
    case 1:
      tree.insert(x, i); 
      break;
    case 2:
      tree.del(x, i);
      break; 
    case 3:
      printf("%d\n", tree.find_rank(x, i)); 
      break; 
    case 4:
      printf("%d\n", tree.find_kth(root[i], x)); 
      break;
    case 5:
      printf("%d\n", tree.get_pre(x, i)); 
      break; 
    case 6:
      printf("%d\n", tree.get_next(x, i));
      break; 
    }
    // printf("root: %d ", root[i]); 
    // tree.dfs(root[i]); printf("\n"); 
    // printf("opt: %d\n", opt); 
  }
  return 0; 
}