You come home and fell some unpleasant smell. Where is it coming from?

You are given an array a. You have to answer the following queries:

1. You are given two integers l
   and r. Let c_i be the number of occurrences of i in a_l: r, where a_l: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c₀, c₁, ..., c_10⁹}
2. You are given two integers p to x. Change a_p to x.

The Mex of a multiset of numbers is the smallest non-negative integer not in the set.

Note that in this problem all elements of a are positive, which means that c₀ = 0 and 0 is never the answer for the query of the second type.

The first line of input contains two integers n and q (1 ≤ n, q ≤ 100 000) — the length of the array and the number of queries respectively.

The second line of input contains n integers — a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Each of the next q lines describes a single query.

The first type of query is described by three integers t_i = 1, l_i, r_i, where 1 ≤ l_i ≤ r_i ≤ n — the bounds of the subarray.

The second type of query is described by three integers t_i = 2, p_i, x_i, where 1 ≤ p_i ≤ n is the index of the element, which must be changed and 1 ≤ x_i ≤ 10⁹ is the new value.

For each query of the first type output a single integer — the Mex of {c₀, c₁, ..., c_10⁹}.

10 4
1 2 3 1 1 2 2 2 9 9
1 1 1
1 2 8
2 7 1
1 2 8

#include<cstdio>
#include<cstring>
#include<map>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=2e5+88;
map<int,int>M;
int vis[N],num[N],a[N],b[N],now[N],ans[N],unit,l,r,t;
struct Query{
    int l,r,tim,id;
    bool operator < (const Query &A)const{
      return l/unit==A.l/unit?(r/unit==A.r/unit?tim<A.tim:r<A.r):l<A.l;
    }
}Q[N];
struct Change{
    int pos,x,y;
}C[N];
void modify(int x,int d){
    --vis[num[x]];num[x]+=d;++vis[num[x]];
}
void going(int T,int d){
    if(C[T].pos>=l&&C[T].pos<=r) modify(C[T].x,d),modify(C[T].y,-d);
    if(d==1) a[C[T].pos]=C[T].x;else a[C[T].pos]=C[T].y;
}
int calc(){
    for(int i=1;;++i) if(!vis[i]) return i;
}
int main(){
    int n,q,tot,op,cc=0,pp=0;
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]),now[i]=b[i]=a[i];
    tot=n,unit=(int)pow(n,0.6666666);
    for(int i=1;i<=q;++i) {
        scanf("%d",&op);
        if(op==2){
        ++cc,scanf("%d%d",&C[cc].pos,&C[cc].x);
        C[cc].y=now[C[cc].pos],b[++tot]=now[C[cc].pos]=C[cc].x;
        }
        else {
        ++pp,scanf("%d%d",&Q[pp].l,&Q[pp].r);
        Q[pp].tim=cc,Q[pp].id=pp;
        }
    }
    sort(b+1,b+tot+1);
    tot=unique(b+1,b+tot+1)-b-1;
    for(int i=1;i<=tot;++i) M[b[i]]=i;
    for(int i=1;i<=n;++i) a[i]=M[a[i]];
    for(int i=1;i<=cc;++i) C[i].x=M[C[i].x],C[i].y=M[C[i].y];
    sort(Q+1,Q+pp+1);
    for(int i=1;i<=pp;++i) {
        while(t<Q[i].tim) going(t+1,1),++t;
        while(t>Q[i].tim) going(t,-1),--t;
        
        while(l<Q[i].l) modify(a[l],-1),++l;
        while(l>Q[i].l) modify(a[l-1],1),--l;
        while(r<Q[i].r) modify(a[r+1],1),++r;
        while(r>Q[i].r) modify(a[r],-1),--r;
        ans[Q[i].id]=calc();
        
    }
    for(int i=1;i<=pp;++i) printf("%d\n",ans[i]);
}

F. Machine Learning 帶修端點莫隊