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 = 1e5 + 10;
const int B = 10;
int a[N], id[N], cnt[2 * N][2], ans[N];
pair<int, int> rep[N];
pair<int, pair<int, int> > qry[N];
vector<int> vec;
bool cmp(int i, int j) {
  return make_pair(qry[i].first >> B,
                   make_pair(qry[i].second.first >> B, qry[i].second.second)) <
         make_pair(qry[j].first >> B,
                   make_pair(qry[j].second.first >> B, qry[j].second.second));
}
void upd(int x, int y) {
  cnt[cnt[x][0]][1]--;
  cnt[x][0] += y;
  cnt[cnt[x][0]][1]++;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  vec.push_back(0);
  for (int i = 1; i <= n; i++) cin >> a[i], vec.push_back(a[i]);
  int c1 = 0, c2 = 0;
  for (int i = 1; i <= q; i++) {
    int typ, x, y;
    cin >> typ >> x >> y;
    if (typ == 1)
      qry[++c1] = make_pair(c2, make_pair(x, y)), id[c1] = c1;
    else
      rep[++c2] = make_pair(x, y), vec.push_back(y);
  }
  sort((vec).begin(), (vec).end());
  vec.resize(unique((vec).begin(), (vec).end()) - vec.begin());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound((vec).begin(), (vec).end(), a[i]) - vec.begin();
  for (int i = 1; i <= c2; i++)
    rep[i].second =
        lower_bound((vec).begin(), (vec).end(), rep[i].second) - vec.begin();
  sort(id + 1, id + c1 + 1, cmp);
  cnt[0][0] = 10 * N;
  int t = 0, l = 1, r = 0;
  for (int i = 1; i <= c1; i++) {
    int I = id[i], T = qry[I].first, L = qry[I].second.first,
        R = qry[I].second.second;
    while (L < l) l--, upd(a[l], 1);
    while (r < R) r++, upd(a[r], 1);
    while (l < L) upd(a[l], -1), l++;
    while (R < r) upd(a[r], -1), r--;
    while (t < T) {
      t++;
      int p = rep[t].first;
      if (L <= p && p <= R) upd(a[p], -1), upd(rep[t].second, 1);
      swap(a[p], rep[t].second);
    }
    while (T < t) {
      int p = rep[t].first;
      if (L <= p && p <= R) upd(a[p], -1), upd(rep[t].second, 1);
      swap(a[p], rep[t].second);
      t--;
    }
    ans[I] = 1;
    while (cnt[ans[I]][1]) ans[I]++;
  }
  for (int i = 1; i <= c1; i++) cout << ans[i] << '\n';
  return 0;
}