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;
template <class _E>
inline void read(_E &e) {
  e = 0;
  bool ck = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ck = 1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    e = (e << 1) + (e << 3) + (ch ^ 48);
    ch = getchar();
  }
  if (ck) e = -e;
}
const int N = 100050;
int pos[N], a[N], sz, n, Q, has, cnt[N], now[N], cntq, vis[N << 1];
map<int, int> mp;
int L = 1, R;
struct querys {
  int t, l, r;
  int id;
} q[N];
struct updata {
  int p, v, y;
} c[N];
inline bool cmp(const querys &x, const querys &y) {
  if (pos[x.l] == pos[y.l] && pos[x.r] == pos[y.r])
    return x.t < y.t;
  else if (pos[x.l] == pos[y.l])
    return pos[x.r] < pos[y.r];
  else
    return pos[x.l] < pos[y.l];
}
int ans[N], tim;
inline void ad(int x) {
  int &y = vis[x];
  if (y > 0) cnt[y]--;
  y++;
  if (y > 0) cnt[y]++;
}
inline void de(int x) {
  int &y = vis[x];
  if (y > 0) cnt[y]--;
  y--;
  if (y > 0) cnt[y]++;
}
inline void addt(int t) {
  int p = c[t].p, v = c[t].v;
  if (L <= p && p <= R) {
    de(a[p]);
    ad(v);
  }
  a[p] = v;
}
inline void delt(int t) {
  int p = c[t].p, v = c[t].y;
  if (L <= p && p <= R) {
    de(a[p]);
    ad(v);
  }
  a[p] = v;
}
int main() {
  read(n), read(Q);
  sz = 2000;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    if (!mp[a[i]]) mp[a[i]] = ++has;
    now[i] = a[i] = mp[a[i]];
    pos[i] = i / sz + 1;
  }
  for (int i = 1; i <= Q; ++i) {
    int opt, x, y;
    read(opt), read(x), read(y);
    if (opt == 1) {
      ++cntq;
      q[cntq] = (querys){tim, x, y, cntq};
    } else {
      if (!mp[y]) mp[y] = ++has;
      y = mp[y];
      ++tim;
      c[tim] = (updata){x, y, now[x]};
      now[x] = y;
    }
  }
  tim = 0;
  sort(q + 1, q + cntq + 1, cmp);
  for (int i = 1; i <= cntq; ++i) {
    while (tim < q[i].t) ++tim, addt(tim);
    while (tim > q[i].t) delt(tim), --tim;
    while (L < q[i].l) de(a[L]), ++L;
    while (R > q[i].r) de(a[R]), --R;
    while (L > q[i].l) --L, ad(a[L]);
    while (R < q[i].r) ++R, ad(a[R]);
    for (int j = 1;; ++j)
      if (!cnt[j]) {
        ans[q[i].id] = j;
        break;
      }
  }
  for (int i = 1; i <= cntq; ++i) printf("%d\n", ans[i]);
  return 0;
}