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
1
2
3
4
5
6
output
2
3
2

1.查询：区间所有数的次数第一个为0的自然数，假如这个区间1出现2次，3出现1次，则答案是3，注意是出现次数

2、修改：修改某一个位置的值。

///带莫队算法模板：求区间不同数的个数+单点修改#include<cstdio>#include<iostream>#include<fstream>#include<algorithm>#include<functional>#include<cstring>#include<string>#include<cstdlib>#include<iomanip>#include<numeric>#include<cctype>#include<cmath>#include<ctime>#include<queue>#include<stack>#include<list>#include<set>#include<map>using namespace std;const int maxn=1000005;typedef long long ll;void add(int x);void del(int x);void modify(int x,int ti);  //这个函数会执行或回退修改ti（执行还是回退取决于是否执行过，具体通过swap实现），x表明当前的询问是x，即若修改了区间[q[x].l,q[x].r]便要更新答案int sz,cnt[maxn],b[maxn],a[maxn],Ans,ans[maxn],num[maxn];struct Change{    int p,col;}c[maxn];struct Query{    int l,r,t,id;    bool operator<(Query& b)    {        return l/sz==b.l/sz?(r/sz==b.r/sz?t<b.t:r<b.r):l<b.l;    }}q[maxn];int main(){     int n,m;        while(scanf("%d%d",&n,&m)!=-1)    {    memset(cnt,0,sizeof(cnt));      memset(num,0,sizeof(num));    sz=pow(n,0.66666);            for (int i=1;i<=n;i++)    {       scanf("%d",&a[i]);       b[i]=a[i];    }    int op;    int ccnt=0,qcnt=0;    for (int i=1;i<=m;++i)    {        scanf("%d",&op);        if (op==1)        {            ++qcnt;            scanf("%d%d",&q[qcnt].l,&q[qcnt].r) ;            q[qcnt].t=ccnt;            q[qcnt].id=qcnt;                    }        else        {            ++ccnt;            scanf("%d%d",&c[ccnt].p,&c[ccnt].col);            b[n+ccnt]=c[ccnt].col;        }    }           sort(b+1,b+n+ccnt+1);    int sum=unique(b+1,b+n+ccnt+1)-b-1;    for(int i=0;i<=n;i++)    {        a[i]=lower_bound(b+1,b+sum+1,a[i])-b;        //cout<<a[i]<<endl;    }    for(int i=1;i<=ccnt;i++)    {        c[i].col=lower_bound(b+1,b+sum+1,c[i].col)-b;    }    int l=1,r=0,now=0;    Ans=0;    sort(q+1,q+qcnt+1);    for (int i=1;i<=qcnt;++i)    {        while (r<q[i].r)        {            add(a[++r]);        }       while(l>q[i].l)        {            add(a[--l]);        }        while (l<q[i].l)        {            del(a[l++]);        }        while (r>q[i].r)        {            del(a[r--]);        }        while (now<q[i].t)        {            modify(i,++now);        }        while (now>q[i].t)        {            modify(i,now--);        }                for(ans[q[i].id]=1;num[ans[q[i].id]]>0;++ans[q[i].id]);            }         for (int i=1;i<=qcnt;++i)     {        cout<<ans[i]<<endl;     }    }        return 0;}void add(int x){    num[cnt[x]]--;    cnt[x]++;    num[cnt[x]]++;}void del(int x){    num[cnt[x]]--;    cnt[x]--;    num[cnt[x]]++;}void modify(int x,int ti){    if (c[ti].p>=q[x].l&&c[ti].p<=q[x].r)    {        del(a[c[ti].p]);        add(c[ti].col);    }    swap(a[c[ti].p],c[ti].col); }