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 + 7, B = 2156;
struct Query {
  int l, r, q, ind;
  Query(int l, int r, int q, int ind) : l(l), r(r), q(q), ind(ind){};
};
struct ChangeQuery {
  int ind, oval, nval;
  ChangeQuery(int ind, int oval, int nval) : ind(ind), oval(oval), nval(nval){};
};
inline bool operator<(const Query &a, const Query &b) {
  if (a.q / B != b.q / B) {
    return a.q < b.q;
  }
  if (a.l / B != b.l / B) {
    return a.l < b.l;
  }
  return a.r < b.r;
}
int n, q, cnt[N], arr[N], actArr[N], ans[N], act[2 * N];
vector<ChangeQuery> changeQueries;
vector<Query> queries;
int actQ = 0;
set<int> allNumbers;
unordered_map<int, int> id;
inline void addVal(int v) {
  cnt[act[v]]--;
  act[v]++;
  cnt[act[v]]++;
}
inline void eraseVal(int v) {
  cnt[act[v]]--;
  act[v]--;
  cnt[act[v]]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    actArr[i] = arr[i];
    allNumbers.insert(arr[i]);
  }
  int actPassed = 0;
  changeQueries.emplace_back(0, 0, 0);
  for (int i = 0; i < q; ++i) {
    int type, a, b;
    cin >> type >> a >> b;
    if (type == 1) {
      a--;
      b--;
      queries.push_back(Query(a, b, actPassed, i - actPassed));
    } else {
      a--;
      changeQueries.push_back(ChangeQuery(a, actArr[a], b));
      actArr[a] = b;
      allNumbers.insert(b);
      actPassed++;
    }
  }
  int actId = 0;
  for (auto i : allNumbers) {
    id[i] = actId++;
  }
  for (int i = 0; i < n; ++i) {
    arr[i] = id[arr[i]];
  }
  for (int i = 0; i < changeQueries.size(); ++i) {
    changeQueries[i].oval = id[changeQueries[i].oval];
    changeQueries[i].nval = id[changeQueries[i].nval];
  }
  sort(queries.begin(), queries.end());
  int l = 0, r = 0;
  cnt[0] = 1e9;
  cnt[1] = 1;
  act[arr[0]] = 1;
  for (int i = 0; i < queries.size(); ++i) {
    int ltrg = queries[i].l, rtrg = queries[i].r, Q = queries[i].q,
        ind = queries[i].ind;
    while (actQ < Q) {
      ++actQ;
      int ind = changeQueries[actQ].ind, oval = changeQueries[actQ].oval,
          nval = changeQueries[actQ].nval;
      if (l <= ind && ind <= r) {
        eraseVal(oval);
        addVal(nval);
      }
      arr[ind] = nval;
    }
    while (actQ > Q) {
      int ind = changeQueries[actQ].ind, oval = changeQueries[actQ].oval,
          nval = changeQueries[actQ].nval;
      --actQ;
      if (l <= ind && ind <= r) {
        eraseVal(nval);
        addVal(oval);
      }
      arr[ind] = oval;
    }
    while (l > ltrg) {
      addVal(arr[--l]);
    }
    while (l < ltrg) {
      eraseVal(arr[l++]);
    }
    while (r < rtrg) {
      addVal(arr[++r]);
    }
    while (r > rtrg) {
      eraseVal(arr[r--]);
    }
    int res = 0;
    while (cnt[res] != 0) {
      res++;
    }
    ans[ind] = res;
  }
  for (int i = 0; i < queries.size(); ++i) {
    cout << ans[i] << "\n";
  }
  return 0;
}