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 int maxn = 2e5 + 5;
int a[maxn], last[maxn], bl[maxn], br[maxn], cnt[maxn], num[maxn], result[maxn];
int size = 0, l = 1, r = 0;
unordered_map<int, int> mp;
struct query {
  int l, r, id, tim;
} Q[maxn];
struct replace {
  int x, c, last;
} R[maxn];
int getid(int x) {
  if (!mp[x]) mp[x] = ++size;
  return mp[x];
}
bool cmp(query a, query b) {
  return bl[a.id] == bl[b.id]
             ? (br[a.id] == br[b.id] ? a.tim < b.tim : br[a.id] < br[b.id])
             : bl[a.id] < bl[b.id];
}
void change(int x, int c) {
  if (x >= l && x <= r) {
    num[cnt[a[x]]]--;
    cnt[a[x]]--;
    num[cnt[a[x]]]++;
    a[x] = c;
    num[cnt[a[x]]]--;
    cnt[a[x]]++;
    num[cnt[a[x]]]++;
  } else
    a[x] = c;
}
void del(int x) {
  if (cnt[a[x]] > 0) num[cnt[a[x]]]--;
  cnt[a[x]]--;
  if (cnt[a[x]] > 0) num[cnt[a[x]]]++;
}
void add(int x) {
  if (cnt[a[x]] > 0) num[cnt[a[x]]]--;
  cnt[a[x]]++;
  if (cnt[a[x]] > 0) num[cnt[a[x]]]++;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  int block = (int)pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] = getid(a[i]);
    last[i] = a[i];
  }
  int opt, k = 0, rd = 0;
  for (int i = 1; i <= q; i++) {
    scanf("%d", &opt);
    if (opt == 1) {
      ++k;
      scanf("%d%d", &Q[k].l, &Q[k].r);
      Q[k].id = k;
      Q[k].tim = rd;
      bl[k] = Q[k].l / block;
      br[k] = Q[k].r / block;
    } else {
      ++rd;
      scanf("%d%d", &R[rd].x, &R[rd].c);
      R[rd].c = getid(R[rd].c);
      R[rd].last = last[R[rd].x];
      last[R[rd].x] = R[rd].c;
    }
  }
  sort(Q + 1, Q + k + 1, cmp);
  int head = 0;
  for (int i = 1; i <= k; i++) {
    while (head < Q[i].tim) {
      head++;
      change(R[head].x, R[head].c);
    }
    while (head > Q[i].tim) {
      change(R[head].x, R[head].last);
      head--;
    }
    while (l < Q[i].l) del(l++);
    while (l > Q[i].l) add(--l);
    while (r < Q[i].r) add(++r);
    while (r > Q[i].r) del(r--);
    for (result[Q[i].id] = 1; num[result[Q[i].id]] > 0; result[Q[i].id]++)
      ;
  }
  for (int i = 1; i <= k; i++) printf("%d\n", result[i]);
  return 0;
}