F. Machine Learning

time limit per test

4 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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 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}
2. 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

Copy

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

Copy

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.

题意：

给你n(n<=1e5)个数，m次查询（m<=1e5），每次查询id x y

如果id=1，则求a[x]~a[y]中所有数出现的次数中，从1开始第一个没出现数。（例如有出现1次、2次、4次的数，则答案为3）

id=2，则把a[x]值改为y

思路：

标准的带修莫队。

首先将a[i]离散化，因为我们要求每个数出现的次数，与值本身无关。

然后贴一个标准的待修莫队的板子。

至于答案，只需要暴力找就可以了，因为最多也就几百次。

有一个奇怪的坑点：如果查询和修改的下标从0开始，本地测试通过，交上去却WA on test 1 ，希望有知道原因的小伙伴可以评论告诉我。。。

代码:

#include <bits/stdc++.h>
using namespace std;
const int maxn=200010;
int n,m,sz;
int a[maxn],now[maxn];
int num[maxn],cnt[maxn],ans[maxn];
int be[maxn],siz;
map<int,int>mp;
struct node
{
    int l,r,tim,id;
    bool operator<(node aa)const
    {
        return be[l]==be[aa.l]?(be[r]==be[aa.r]?tim<aa.tim:r<aa.r):l<aa.l;
    }
}q[maxn];
int m1,m2;
struct Change
{
    int pos,nw,od;
}c[maxn];

int l,r,t;
int getid(int x)
{
    if(!mp.count(x)) mp[x]=++siz;
    return mp[x];
}
void add(int i,int v)
{
    if(num[i]>0) cnt[num[i]]--;
    num[i]+=v;
    if(num[i]>0) cnt[num[i]]++;
}
void go(int i,int d)
{
    if(l<=i&&i<=r)
    {
        add(a[i],-1);
        add(d,1);
    }
    a[i]=d;
}

int main()
{
    int T,cas=1;
    while(scanf("%d%d",&n,&m)!=EOF)
    //scanf("%d%d",&n,&m);
    {
        sz=pow(n,0.666666);
        memset(num,0,sizeof(num));
        memset(cnt,0,sizeof(cnt));
        mp.clear();
        siz=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            now[i]=a[i]=getid(a[i]);
            be[i]=(i-1)/sz+1;
        }
        m1=0;m2=0;
        for(int i=0;i<m;i++)
        {
            int op,x,y;
            scanf("%d%d%d",&op,&x,&y);
            if(op==1) q[++m1]=(node){x,y,m2,m1};
            else
            {
                y=getid(y);
                c[++m2]=(Change){x,y,now[x]};
                now[x]=y;
            }
        }
        sort(q+1,q+m1+1);
        l=1;r=0;t=0;
        for(int i=1;i<=m1;i++)
        {
            while(t<q[i].tim) {t++;go(c[t].pos,c[t].nw);}
            while(t>q[i].tim) {go(c[t].pos,c[t].od);t--;}

            while(l<q[i].l){add(a[l],-1);l++;}
            while(l>q[i].l){add(a[l-1],1);l--;}
            while(r<q[i].r){add(a[r+1],1);r++;}
            while(r>q[i].r){add(a[r],-1);r--;}

            int x=1;
            while(cnt[x]>0) x++;
            ans[q[i].id]=x;
        }
        for(int i=1;i<=m1;i++)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}