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 = 200005, K = 3005;
struct node {
  int l, r, tim, id;
} q[N];
struct node2 {
  int v, id, kd;
} ls[N];
struct node3 {
  int p, v;
} q2[N];
int n, m, nn, q1, qt, now, opt, size, size2, bl[N], a[N], ans[N], tot[N],
    totk[K], cnt[N], bl2[N];
bool cmd(node a, node b) {
  if (bl[a.l] < bl[b.l]) return true;
  if (bl[a.l] > bl[b.l]) return false;
  if (bl[a.r] < bl[b.r]) return true;
  if (bl[a.r] > bl[b.r]) return false;
  return a.tim < b.tim;
}
bool cmd2(node2 a, node2 b) { return a.v < b.v; }
void add(int k) {
  if (--tot[cnt[k]] == 0) --totk[bl2[cnt[k]]];
  if (tot[++cnt[k]]++ == 0) ++totk[bl2[cnt[k]]];
}
void del(int k) {
  if (--tot[cnt[k]] == 0) --totk[bl2[cnt[k]]];
  if (tot[--cnt[k]]++ == 0) ++totk[bl2[cnt[k]]];
}
int getans() {
  for (int i = 1; i <= bl2[now]; ++i)
    if (totk[i] != size2)
      for (int j = (i - 1) * size2 + 1; j <= min(now, i * size2); ++j)
        if (!tot[j]) return j;
}
int main() {
  scanf("%d%d", &n, &m);
  nn = n;
  if (n <= 5000)
    size = pow(n, 0.57);
  else
    size = pow(n, 0.73);
  if (!size) size = 1;
  for (int i = 1; i <= n; ++i)
    scanf("%d", &ls[i].v), ls[i].id = i, ls[i].kd = 1,
                           bl[i] = (i - 1) / size + 1;
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &opt);
    if (opt == 1)
      ++q1, scanf("%d%d", &q[q1].l, &q[q1].r), q[q1].tim = qt, q[q1].id = q1;
    else
      ++qt, scanf("%d%d", &q2[qt].p, &q2[qt].v), ls[++nn].v = q2[qt].v,
                                                 ls[nn].id = qt, ls[nn].kd = 2;
  }
  if (q1) sort(q + 1, q + 1 + q1, cmd);
  if (nn) sort(ls + 1, ls + 1 + nn, cmd2);
  for (int i = 1; i <= nn; ++i) {
    if (ls[i].v != ls[i - 1].v) ++now;
    if (ls[i].kd == 1)
      a[ls[i].id] = now;
    else
      q2[ls[i].id].v = now;
  }
  size2 = sqrt(now);
  if (!size2) size2 = 1;
  for (int i = 1; i <= now; ++i) bl2[i] = (i - 1) / size2 + 1;
  for (int l = 1, r = 0, tim = 0, i = 1; i <= q1; ++i) {
    while (r < q[i].r) add(a[++r]);
    while (l > q[i].l) add(a[--l]);
    while (r > q[i].r) del(a[r--]);
    while (l < q[i].l) del(a[l++]);
    while (tim < q[i].tim) {
      ++tim;
      if (l <= q2[tim].p && q2[tim].p <= r) {
        del(a[q2[tim].p]);
        swap(q2[tim].v, a[q2[tim].p]);
        add(a[q2[tim].p]);
      } else
        swap(q2[tim].v, a[q2[tim].p]);
    }
    while (tim > q[i].tim) {
      if (l <= q2[tim].p && q2[tim].p <= r) {
        del(a[q2[tim].p]);
        swap(q2[tim].v, a[q2[tim].p]);
        add(a[q2[tim].p]);
      } else
        swap(q2[tim].v, a[q2[tim].p]);
      --tim;
    }
    ans[q[i].id] = getans();
  }
  for (int i = 1; i <= q1; ++i) printf("%d\n", ans[i]);
}