1、题目大意：给你一个序列，有5种操作，都有什么呢。。。

1> 区间第k小 这个直接用二分+树套树做

2> 区间小于k的有多少 这个直接用树套树做

3> 单点修改 这个直接用树套树做

4> 区间内k的前驱 这个就是1和2操作的合并，就是查询k的排名，然后就是知道他的前驱的排名，然后第k小

5> 区间内k的后继 这个和4同理

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
struct Node{
    Node *ch[2];
    int num, cnt, v, r;
    bool operator < (const Node& rhs) const{
        return r < rhs.r;
    }
    inline int cmp(int x){
        if(x == v) return -1;
        if(x < v) return 0;
        return 1;
    }
    inline void maintain(){
        cnt = num;
        if(ch[0]) cnt += ch[0] -> cnt;
        if(ch[1]) cnt += ch[1] -> cnt;
        return;
    }
} ft[5000000], *root[400000];
int tot;
int a[100000];
inline void treap_rotate(Node* &o, int d){
    Node *k = o -> ch[d ^ 1];
    o -> ch[d ^ 1] = k  -> ch[d];
    k -> ch[d] = o;
    o -> maintain();
    k -> maintain();
    o = k;
    return;
}
inline void treap_insert(Node* &o, int value){
    if(!o){
        o = &ft[tot ++];
        o -> ch[0] = o -> ch[1] = NULL;
        o -> num = 1;
        o -> v = value;
        o -> r = rand();
    }
    else{
        int d = o -> cmp(value);
        if(d == -1){
            o -> num ++;
        }
        else{
            treap_insert(o -> ch[d], value);
            if(o -> ch[d] > o) treap_rotate(o, d ^ 1);
        }
    }
    o -> maintain();
}
inline void treap_remove(Node* &o, int value){
    if(!o) return;
    int d = o -> cmp(value);
    if(d == -1){
        if(o -> num > 1) o -> num --;
        else if(!o -> ch[0]) o = o -> ch[1];
        else if(!o -> ch[1]) o = o -> ch[0];
        else{
            int d2;
            if(o -> ch[0] > o -> ch[1]) d2 = 1;
            else d2 = 0;
            treap_rotate(o, d2);
            treap_remove(o -> ch[d2], value);
        }
    }
    else{
        treap_remove(o -> ch[d], value);
    }
    if(o) o -> maintain();
}
inline int treap_lessk(Node* &o, int k){
    if(!o) return 0;
    int d = o -> cmp(k);
    if(d == -1){
        int ret = o -> num;
        if(o -> ch[0]) ret += o -> ch[0] -> cnt;
        return ret;
    }
    else if(d == 0){
        return treap_lessk(o -> ch[d], k);
    }
    else{
        int ss = o -> num;
        if(o -> ch[0]) ss += o -> ch[0] -> cnt;
        return treap_lessk(o -> ch[d], k) + ss;
    }
}
inline int treap_find(Node* &o, int k){
    if(!o) return 0;
    int d = o -> cmp(k);
    if(d == -1) return o -> num;
    else return treap_find(o -> ch[d], k);
}
inline int segment_tree_find(int l, int r, int o, int x, int y, int k){
    if(x <= l && r <= y){
        return treap_find(root[o], k);
    }
    int mid = (l + r) / 2;
    int ret = 0;
    if(x <= mid) ret += segment_tree_find(l, mid, 2 * o, x, y, k);
    if(y > mid) ret += segment_tree_find(mid + 1, r, 2 * o + 1, x, y, k);
    return ret;
}
inline void segment_tree_add(int l, int r, int o, int x, int value){
    treap_remove(root[o], a[x]);
    treap_insert(root[o], value);
    if(l == r){
        a[x] = value;
        return;
    }
    int mid = (l + r) / 2;
    if(x <= mid) segment_tree_add(l, mid, 2 * o, x, value);
    else segment_tree_add(mid + 1, r, 2 * o + 1, x, value);
}
inline int segment_tree_query_lessk(int l, int r, int o, int x, int y, int k){
    if(x <= l && r <= y){
        return treap_lessk(root[o], k);
    }
    int mid = (l + r) / 2;
    int ret = 0;
    if(x <= mid) ret += segment_tree_query_lessk(l, mid, 2 * o, x, y, k);
    if(y > mid) ret += segment_tree_query_lessk(mid + 1, r, 2 * o + 1, x, y, k);
    return ret;
}
inline int segment_tree_query_NO_k(int l, int r, int o, int x, int y, int k, int n){
    int L = 0, R = 100000000;
    while(L < R){
        int mid = (L + R) / 2;
        if(segment_tree_query_lessk(1, n, 1, x, y, mid) < k) L = mid + 1;
        else R = mid;
    }
    return L;
}
int main(){
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++){
        int x;
        scanf("%d", &x);
        segment_tree_add(1, n, 1, i, x);
    }
    for(int i = 1; i <= m; i ++){
        int op, x, y, z;
        scanf("%d", &op);
        if(op == 1){
            scanf("%d%d%d", &x, &y, &z);
            int ans = segment_tree_query_lessk(1, n, 1, x, y, z);
            ans -= segment_tree_find(1, n, 1, x, y, z);
            ans ++;
            printf("%d\n", ans);
        }
        else if(op == 2){
            scanf("%d%d%d", &x, &y, &z);
            int ans = segment_tree_query_NO_k(1, n, 1, x, y, z, n);
            printf("%d\n", ans);
        }
        else if(op == 3){
            scanf("%d%d", &x, &y);
            segment_tree_add(1, n, 1, x, y);
        }
        else if(op == 4){
            scanf("%d%d%d", &x, &y, &z);
            int k = segment_tree_query_lessk(1, n, 1, x, y, z);
            k -= segment_tree_find(1, n, 1, x, y, z);
            int ans = segment_tree_query_NO_k(1, n, 1, x, y, k, n);
            printf("%d\n", ans);
        }
        else{
            scanf("%d%d%d", &x, &y, &z);
            int k = segment_tree_query_lessk(1, n, 1, x, y, z);
            k ++;
            int ans = segment_tree_query_NO_k(1, n, 1, x, y, k, n);
            printf("%d\n", ans);
        }
    }
    return 0;
}