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;
int n, m;
int a[200005], a1[200005], ans[200005], cnt[200005], cnt2[200005];
map<int, int> num;
struct node {
  int l, r, x, id;
  bool operator<(const node &a) const {
    if (l / 2000 != a.l / 2000) return l / 2000 < a.l / 2000;
    if (r / 2000 != a.r / 2000) return r / 2000 < a.r / 2000;
    return x < a.x;
  }
} query[200005];
struct change {
  int pos, val, rval;
} ch[200005];
void add(int val) {
  if (cnt[val] > 0) cnt2[cnt[val]]--;
  cnt[val]++;
  cnt2[cnt[val]]++;
}
void remov(int val) {
  cnt2[cnt[val]]--;
  cnt[val]--;
  if (cnt[val] > 0) cnt2[cnt[val]]++;
}
void work(int x, int i) {
  if (ch[x].pos >= query[i].l && ch[x].pos <= query[i].r) {
    remov(ch[x].val);
    add(ch[x].rval);
  }
  a[ch[x].pos] = ch[x].rval;
}
void rework(int x, int i) {
  if (ch[x].pos >= query[i].l && ch[x].pos <= query[i].r) {
    remov(ch[x].rval);
    add(ch[x].val);
  }
  a[ch[x].pos] = ch[x].val;
}
int main() {
  int ncnt = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!num[a[i]]) num[a[i]] = ++ncnt;
    a[i] = num[a[i]];
    a1[i] = a[i];
  }
  int now = 0, q = 0;
  for (int i = 1; i <= m; i++) {
    int op, l, r;
    scanf("%d%d%d", &op, &l, &r);
    if (op == 2) {
      if (!num[r]) num[r] = ++ncnt;
      r = num[r];
      ch[++now].val = a1[l];
      ch[now].pos = l;
      ch[now].rval = r;
      a1[l] = r;
    } else {
      query[++q].l = l;
      query[q].r = r;
      query[q].x = now;
      query[q].id = q;
    }
  }
  sort(query + 1, query + q + 1);
  int l = 1, r = 1;
  now = 0;
  add(a[1]);
  for (int i = 1; i <= q; i++) {
    int L = query[i].l, R = query[i].r, time = query[i].x, id = query[i].id;
    while (l > L) add(a[--l]);
    while (r < R) add(a[++r]);
    while (l < L) remov(a[l++]);
    while (R < r) remov(a[r--]);
    while (now < time) work(++now, i);
    while (now > time) rework(now--, i);
    for (int j = 1;; j++)
      if (!cnt2[j]) {
        ans[id] = j;
        break;
      }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}