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 long long INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const int M = 1e6 + 10;
const long long mod = 1e9 + 7;
const double PI = acos(-1);
int sq = 3000;
int cntq, cntc;
map<int, int> mp;
int vis[N];
int cnt[N];
int num[N];
int ans[N];
struct query {
  int l, r, z, id;
} qq[N];
struct change {
  int x, y;
} ch[N];
bool cmp1(query a, query b) {
  if (a.l / sq == b.l / sq) {
    if (a.r / sq == b.r / sq)
      return a.z < b.z;
    else
      return a.r < b.r;
  } else
    return a.l < b.l;
}
void update(int x, int y) {
  vis[cnt[num[x]]]--;
  cnt[num[x]] += y;
  vis[cnt[num[x]]]++;
}
void solve() {
  int l = 1, r = 1, c = 0, res;
  update(1, 1);
  for (int i = 1; i <= cntq; ++i) {
    while (r < qq[i].r) update(++r, 1);
    while (r > qq[i].r) update(r--, -1);
    while (l > qq[i].l) update(--l, 1);
    while (l < qq[i].l) update(l++, -1);
    while (c > qq[i].z) {
      if (ch[c].x >= l && ch[c].x <= r) update(ch[c].x, -1);
      swap(ch[c].y, num[ch[c].x]);
      if (ch[c].x >= l && ch[c].x <= r) update(ch[c].x, 1);
      c--;
    }
    while (c < qq[i].z) {
      c++;
      if (ch[c].x >= l && ch[c].x <= r) update(ch[c].x, -1);
      swap(ch[c].y, num[ch[c].x]);
      if (ch[c].x >= l && ch[c].x <= r) update(ch[c].x, 1);
    }
    res = 1;
    while (vis[res]) res++;
    ans[qq[i].id] = res;
  }
}
int main() {
  int n, q, tot = 0, x, op, l, r;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &x);
    if (mp[x])
      num[i] = mp[x];
    else
      num[i] = mp[x] = ++tot;
  }
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1) {
      ++cntq;
      qq[cntq] = {l, r, cntc, cntq};
    } else {
      ++cntc;
      if (mp[r])
        r = mp[r];
      else
        r = mp[r] = ++tot;
      ch[cntc] = {l, r};
    }
  }
  sort(qq + 1, qq + cntq + 1, cmp1);
  solve();
  for (int i = 1; i <= cntq; ++i) printf("%d\n", ans[i]);
  return 0;
}