Problem: 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

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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int sz = 2000;
int n, q, cntz, cntq, a[101000], vis[101000], now = 0, L = 1, R = 0,
                                              cnt[301000];
vector<int> oo;
struct CZ {
  int p, fr, to;
} cz[101000];
struct QUE {
  int l, r, tim, ans, id;
} que[101000];
void add(int x) {
  --vis[cnt[x]];
  ++vis[++cnt[x]];
}
void del(int x) {
  --vis[cnt[x]];
  ++vis[--cnt[x]];
}
void bak(CZ &t) {
  if (t.p >= L && t.p <= R) {
    del(t.to);
    add(t.fr);
  }
  a[t.p] = t.fr;
}
void fro(CZ &t) {
  if (t.p >= L && t.p <= R) {
    del(t.fr);
    add(t.to);
  }
  a[t.p] = t.to;
}
bool cmp(QUE &x, QUE &y) {
  if (((x.l - 1) / sz) != ((y.l - 1) / sz)) return x.l < y.l;
  if (((x.r - 1) / sz) != ((y.r - 1) / sz)) return x.r < y.r;
  return x.tim < y.tim;
}
bool org(QUE &x, QUE &y) { return x.id < y.id; }
int main() {
  vis[0] = 1e8;
  cin >> n >> q;
  int t, l, r, p, x;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), oo.push_back(a[i]);
  for (int i = 1; i <= q; ++i) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &l, &r);
      que[++cntq] = (QUE){l, r, now, 0, i};
    } else {
      scanf("%d%d", &p, &x);
      oo.push_back(x);
      cz[++cntz] = (CZ){p, a[p], x};
      a[p] = x;
      ++now;
    }
  }
  sort(oo.begin(), oo.end());
  int ss = unique(oo.begin(), oo.end()) - oo.begin();
  oo.resize(ss);
  for (int i = 1; i <= n; ++i)
    a[i] = lower_bound(oo.begin(), oo.end(), a[i]) - oo.begin() + 1;
  for (int i = 1; i <= cntz; ++i) {
    cz[i].fr = lower_bound(oo.begin(), oo.end(), cz[i].fr) - oo.begin() + 1;
    cz[i].to = lower_bound(oo.begin(), oo.end(), cz[i].to) - oo.begin() + 1;
  }
  sort(que + 1, que + cntq + 1, cmp);
  for (int i = 1; i <= cntq; ++i) {
    QUE the = que[i];
    for (; R < the.r;) add(a[++R]);
    for (; L > the.l;) add(a[--L]);
    for (; R > the.r;) del(a[R--]);
    for (; L < the.l;) del(a[L++]);
    for (; now < the.tim;) fro(cz[++now]);
    for (; now > the.tim;) bak(cz[now--]);
    for (; vis[que[i].ans];) ++que[i].ans;
  }
  sort(que + 1, que + cntq + 1, org);
  for (int i = 1; i <= cntq; ++i) printf("%d\n", que[i].ans);
}