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 N = 2e6;
vector<int> v;
int n, a[N], cnt[N], app[N], m, bls, bel[N], qcnt, ccnt, ans[N];
struct query {
  int l, r, pre, pos;
  bool operator<(query x) const {
    if (bel[l] == bel[x.l]) {
      if (bel[r] == bel[x.r])
        return pre < x.pre;
      else
        return r < x.r;
    } else
      return l < x.l;
  }
} ask[N];
struct modify {
  int val, pos;
} change[N];
inline int getid(int x) {
  return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}
inline void add(int x) {
  --app[cnt[x]++];
  app[cnt[x]]++;
}
inline void sub(int x) {
  --app[cnt[x]--];
  app[cnt[x]]++;
}
inline void cha(int x, int i) {
  if (change[x].pos >= ask[i].l && change[x].pos <= ask[i].r) {
    sub(a[change[x].pos]);
    add(change[x].val);
  }
  swap(a[change[x].pos], change[x].val);
}
int main() {
  scanf("%d%d", &n, &m);
  bls = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    v.push_back(a[i]);
    bel[i] = i / bls;
  }
  int opt;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &opt);
    if (opt == 1) {
      qcnt++;
      scanf("%d%d", &ask[qcnt].l, &ask[qcnt].r);
      ask[qcnt].pos = qcnt;
      ask[qcnt].pre = ccnt;
    } else {
      ccnt++;
      scanf("%d%d", &change[ccnt].pos, &change[ccnt].val);
      v.push_back(change[ccnt].val);
    }
  }
  sort(ask + 1, ask + 1 + qcnt);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 1; i <= n; i++) a[i] = getid(a[i]);
  for (int i = 1; i <= ccnt; i++) change[i].val = getid(change[i].val);
  int l = 1, r = 0, now = 0;
  for (int i = 1; i <= qcnt; i++) {
    while (l > ask[i].l) add(a[--l]);
    while (r < ask[i].r) add(a[++r]);
    while (l < ask[i].l) sub(a[l++]);
    while (r > ask[i].r) sub(a[r--]);
    while (now < ask[i].pre) cha(++now, i);
    while (now > ask[i].pre) cha(now--, i);
    for (ans[ask[i].pos] = 1; app[ans[ask[i].pos]]; ans[ask[i].pos]++)
      ;
  }
  for (int i = 1; i <= qcnt; i++) printf("%d\n", ans[i]);
  return 0;
}