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 MAXN = 200005, block = 2000;
int a[MAXN], b[MAXN], ans[MAXN], cnt1[MAXN], cnt2[MAXN], blo[MAXN];
int n, m, tot, AT = 0, mc;
map<int, int> mp;
struct node {
  int l, r, t, id;
} q[MAXN];
struct time {
  int pos, val, rev;
} Ti[MAXN];
int cmp(node a, node b) {
  if (blo[a.l] != blo[b.l]) return blo[a.l] < blo[b.l];
  if (blo[a.r] != blo[b.r]) return blo[a.r] < blo[b.r];
  return a.t < b.t;
}
void add(int now) {
  if (cnt1[now] > 0) cnt2[cnt1[now]]--;
  cnt1[now]++;
  cnt2[cnt1[now]]++;
}
void remove(int now) {
  cnt2[cnt1[now]]--;
  cnt1[now]--;
  if (cnt1[now] > 0) cnt2[cnt1[now]]++;
}
void gotime(int now, int i) {
  if (q[i].l <= Ti[now].pos && Ti[now].pos <= q[i].r) {
    remove(Ti[now].rev);
    add(Ti[now].val);
  }
  a[Ti[now].pos] = Ti[now].val;
}
void retime(int now, int i) {
  if (q[i].l <= Ti[now].pos && Ti[now].pos <= q[i].r) {
    remove(Ti[now].val);
    add(Ti[now].rev);
  }
  a[Ti[now].pos] = Ti[now].rev;
}
int main() {
  scanf("%d %d", &n, &m);
  mc = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (mp[a[i]] == 0) mp[a[i]] = ++mc;
    a[i] = mp[a[i]];
    b[i] = a[i];
    blo[i] = (i - 1) / block + 1;
  }
  tot = 0, AT = 0;
  for (int i = 1; i <= m; i++) {
    int t, l, r;
    scanf("%d %d %d", &t, &l, &r);
    if (t == 1) {
      q[++tot].l = l;
      q[tot].r = r;
      q[tot].t = AT;
      q[tot].id = tot;
    } else {
      Ti[++AT].pos = l;
      if (mp[r] == 0) mp[r] = ++mc;
      r = mp[r];
      Ti[AT].val = r;
      Ti[AT].rev = b[l];
      b[l] = r;
    }
  }
  sort(q + 1, q + tot + 1, cmp);
  int L = 1, R = 1, T = 0;
  add(a[1]);
  for (int i = 1; i <= tot; i++) {
    while (L > q[i].l) add(a[--L]);
    while (R < q[i].r) add(a[++R]);
    while (L < q[i].l) remove(a[L++]);
    while (R > q[i].r) remove(a[R--]);
    while (T < q[i].t) gotime(++T, i);
    while (T > q[i].t) retime(T--, i);
    for (int j = 1;; j++) {
      if (cnt2[j] == 0) {
        ans[q[i].id] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= tot; i++) printf("%d\n", ans[i]);
}