線段樹套平衡樹是什麼腦殘東西，複雜度就是假的，O(nlog3n)$O(nlo{g}^{3}n)$讓人感覺非常不靠譜。所以我們為什麼不用更好寫的樹狀陣列代替線段樹，更好寫的主席樹（權值線段樹）代替平衡樹呢？而且，不僅是好寫，複雜度也是很對的O(nlog2n)$O(nlo{g}^{2}n)$啊。

我們來簡單理解一下樹套樹是什麼：
思想其實很簡單，樹狀陣列的每個節點都是一顆權值線段樹，並且每顆權值線段樹維護的資訊都是樹狀陣列式的累加，這樣每次查詢和修改都只需要對logn個線段樹進行操作。

對比一下普通的樹狀陣列和樹套樹：

額，，，差不多就是這樣了，確實很簡單吧。

下面這個程式碼是二逼平衡樹的板子，除了修改和查詢k大還有求rank和前驅後繼（所以我在講平衡樹的時候說過維護權值的事線段樹也能做嘛）

Code

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>

const int maxn = 5e4 + 7;
const int inf = 0x7fffffff;

using namespace std;

int n, m;
int a[maxn];
int val[maxn << 1], tot;
int opt[maxn];
int qa[maxn];
int qb[maxn];
int qc[maxn];

struct
 
 node {
    int sum;
    int l, r;
} st[maxn * 400];
int cnt;
int root[maxn];

int xx[20], cnt1;
int yy[20], cnt2;

inline int read()
{
    int X = 0; char ch = getchar();
    while (ch < '0' || ch > '9') ch = getchar();
    while (ch >= '0' && ch <= '9') X = X * 10 + ch - '0', ch = getchar();
    return
 
 X;
}

inline int lowbit(int x)
{
    return x & -x;
}

void update(int num, int &rt, int l, int r, int x)
{
    st[++cnt] = st[rt];
    rt = cnt;
    st[rt].sum += x;
    if (l == r) return;
    int mid = l + r >> 1;
    if (num <= mid) update(num, st[rt].l, l, mid, x);
    else update(num, st[rt].r, mid + 1, r, x);
}

int get_rnk(int l, int r, int x)
{
    if (l == r) return 0;
    int mid = l + r >> 1;
    if (x <= mid) {
        for (int i = 1; i <= cnt1; i++) xx[i] = st[xx[i]].l;
        for (int i = 1; i <= cnt2; i++) yy[i] = st[yy[i]].l;
        return get_rnk(l, mid, x);
    }
    int d = 0;
    for (int i = 1; i <= cnt1; i++) d -= st[st[xx[i]].l].sum;
    for (int i = 1; i <= cnt2; i++) d += st[st[yy[i]].l].sum;
    for (int i = 1; i <= cnt1; i++) xx[i] = st[xx[i]].r;
    for (int i = 1; i <= cnt2; i++) yy[i] = st[yy[i]].r;
    return get_rnk(mid + 1, r, x) + d;
}

int get_kth(int l, int r, int k)
{
    if (l == r) return val[l];
    int d = 0, mid = l + r >> 1;
    for (int i = 1; i <= cnt1; i++) d -= st[st[xx[i]].l].sum;
    for (int i = 1; i <= cnt2; i++) d += st[st[yy[i]].l].sum;
    if (k <= d) {
        for (int i = 1; i <= cnt1; i++) xx[i] = st[xx[i]].l;
        for (int i = 1; i <= cnt2; i++) yy[i] = st[yy[i]].l;
        return get_kth(l, mid, k);
    }
    for (int i = 1; i <= cnt1; i++) xx[i] = st[xx[i]].r;
    for (int i = 1; i <= cnt2; i++) yy[i] = st[yy[i]].r;
    return get_kth(mid + 1, r, k - d);
}

inline void add(int i, int x)
{
    int k = lower_bound(val + 1, val + tot + 1, a[i]) - val;
    for (; i <= n; i += lowbit(i)) update(k, root[i], 1, tot, x);
}

inline void init_query(int i)
{
    cnt1 = cnt2 = 0;
    for (int j = qa[i] - 1; j; j -= lowbit(j)) xx[++cnt1] = root[j];
    for (int j = qb[i]; j; j -= lowbit(j)) yy[++cnt2] = root[j];
}

int main(void)
{
    cin >> n >> m;
    tot = n;
    for (int i = 1; i <= n; i++) a[i] = val[i] = read();
    for (int i = 1; i <= m; i++) {
        opt[i] = read();
        qa[i] = read();
        qb[i] = read();
        if (opt[i] != 3) {
            qc[i] = read();
            if (opt[i] != 2) val[++tot] = qc[i];
        }
        else val[++tot] = qb[i];
    }
    sort(val + 1, val + tot + 1);
    tot = unique(val + 1, val + tot + 1) - val - 1;
    st[0] = {0, 0, 0};
    for (int i = 1; i <= n; i++) add(i, 1);
    for (int i = 1; i <= m; i++) {
        if (opt[i] != 3) init_query(i);
        if (opt[i] != 2 && opt[i] != 3) qc[i] = lower_bound(val + 1, val + tot + 1, qc[i]) - val;
        if (opt[i] == 1) printf("%d\n", get_rnk(1, tot, qc[i]) + 1);
        else if (opt[i] == 2) printf("%d\n", get_kth(1, tot, qc[i]));
        else if (opt[i] == 3) {
            add(qa[i], -1);
            a[qa[i]] = qb[i];
            add(qa[i], 1);
        }
        else if (opt[i] == 4) {
            int k = get_rnk(1, tot, qc[i]);
            if (!k) printf("%d\n", -inf);
            else {
                init_query(i);
                printf("%d\n", get_kth(1, tot, k));
            }
        }
        else {
            int k = get_rnk(1, tot, qc[i] + 1);
            if (k > qb[i] - qa[i]) printf("%d\n", inf);
            else {
                init_query(i);
                printf("%d\n", get_kth(1, tot, k + 1));
            }
        }
    }

    return 0;
}