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:

You are given two integers l and r. Let ci be the number of occurrences of i in al: r, where al: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c0, c1, …, c109} You are given two integers p to x. Change ap 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.

Input 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, …, an (1 ≤ ai ≤ 109).

Each of the next q lines describes a single query.

The first type of query is described by three integers ti = 1, li, ri, where 1 ≤ li ≤ ri ≤ n — the bounds of the subarray.

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

Output For each query of the first type output a single integer — the Mex of {c0, c1, …, c109}.

Example Input 10 4 1 2 3 1 1 2 2 2 9 9 1 1 1 1 2 8 2 7 1 1 2 8 Output 2 3 2 Note The subarray of the first query consists of the single element — 1.

The subarray of the second query consists of four 2s, one 3 and two 1s.

The subarray of the fourth query consists of three 1s, three 2s and one 3.

題意：

給你n個數，有兩個操作，1 l r表示在l到r這段區間，詢問{c0,c1,c2,⋯,c109}的Mex，而ci表示數值i在l r中的出現次數，Mex代表未出現的最小的正整數。舉個例子:1 2 3 4 3 2 2 ，最小為4，1 2 3 3 2 1最小是1.。。。2 p x 代表把a[p]改成x

題解：

帶修莫隊，cnt1[x]維護x出現的次數cnt2[x]維護出現x次的數有多少個。把所有2的操作記錄下來，在記錄tim代表這個1操作之前有多少2操作，如果在莫隊的時候的t小了，那就增加2操作，反之減少

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int len;
struct node
{
    int l,r,t,bl,br,indx;
    node(){}
    node(int l,int r,int t,int indx):l(l),r(r),t(t),indx(indx){
        bl=l/len,br=r/len;
    }
    bool operator< (const node& a)const
    {
        if(bl==a.bl&&br==a.br)
            return t<a.t;
        if(bl==a.bl)
            return br<a.br;
        return bl<a.bl;
    }
}que[N];
struct query
{
    int pos,pre,val;// 位置，上一個數，當前的數
    query(){}
    query(int pos,int pre,int val):pos(pos),pre(pre),val(val){}
}c[N];
int cnt1[2*N],cnt2[2*N],a[N],b[N*2],now[N],cnt,ans[N];
void add(int pos)
{
    cnt2[cnt1[pos]]--;
    cnt1[pos]++;
    cnt2[cnt1[pos]]++;
}
void del(int pos)
{
    cnt2[cnt1[pos]]--;
    cnt1[pos]--;
    cnt2[cnt1[pos]]++;
}
int ll,rr;
void change(int pos,int x)
{
    if(pos>=ll&&pos<=rr){
        del(now[pos]);
        add(x);
    }
    now[pos]=x;
}
int main()
{
    int n,q;
    scanf("%d%d",&n,&q);
    len=pow(n,0.666667);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        now[i]=a[i];
        b[++cnt]=a[i];
    }
    /*for(int i=1;i<=n;i++)
        cout<<now[i]<<" ";
    cout<<endl;*/
    int op,qnum=0,tim=0;
    for(int i=1;i<=q;i++)
    {
        scanf("%d%d%d",&op,&ll,&rr);
        if(op==1)
            que[++qnum]=node(ll,rr,tim,qnum);
        else
            b[++cnt]=rr,c[++tim]=query(ll,now[ll],rr),now[ll]=rr;
    }
    /*for(int i=1;i<=n;i++)
        cout<<now[i]<<" ";
    cout<<endl;*/
    sort(b+1,b+1+cnt);
    sort(que+1,que+qnum+1);
    int all=unique(b+1,b+1+cnt)-b-1;
    for(int i=1;i<=n;i++)
        now[i]=lower_bound(b+1,b+1+all,a[i])-b;
    for(int i=1;i<=tim;i++)
    {
        c[i].pre=lower_bound(b+1,b+1+all,c[i].pre)-b;
        c[i].val=lower_bound(b+1,b+1+all,c[i].val)-b;
    }
    ll=1,rr=0;
    int t=0;
    for(int i=1;i<=qnum;i++)
    {
        while(ll>que[i].l)
            ll--,add(now[ll]);
        while(rr<que[i].r)
            rr++,add(now[rr]);
        while(ll<que[i].l)
            del(now[ll]),ll++;
        while(rr>que[i].r)
            del(now[rr]),rr--;
        while(t<que[i].t)
            t++,change(c[t].pos,c[t].val);
        while(t>que[i].t)
            change(c[t].pos,c[t].pre),t--;
        int mex=1;
        while(cnt2[mex])
            mex++;
        ans[que[i].indx]=mex;
    }
    for(int i=1;i<=qnum;i++)
        printf("%d\n",ans[i]);
    return 0;
}