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 = 2e5 + 10;
int n, q, cntq, cntu, cntjs, a[N], cnt[N], js[N], ans[N], belong[N], tot[N];
struct pos {
  int id, l, r, t;
} ask[N];
struct node {
  int x, val;
} upd[N];
bool cmp(pos x, pos y) {
  if (belong[x.l] ^ belong[y.l]) return belong[x.l] < belong[y.l];
  if (belong[x.r] ^ belong[y.r]) return belong[x.r] < belong[y.r];
  if (belong[x.l] & 1) return x.t < y.t;
  return x.t > y.t;
}
void init() {
  sort(js + 1, js + cntjs + 1);
  int len = unique(js + 1, js + cntjs + 1) - js - 1;
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(js + 1, js + len + 1, a[i]) - js;
  int sz = pow(n, 2.0 / 3.0), num = ceil((double)n / sz);
  for (int i = 1; i <= num; i++) {
    int tmp = min(n, i * sz);
    for (int j = (i - 1) * sz + 1; j <= tmp; j++) belong[j] = i;
  }
  for (int i = 1; i <= cntu; i++)
    upd[i].val = lower_bound(js + 1, js + len + 1, upd[i].val) - js;
  sort(ask + 1, ask + cntq + 1, cmp);
  tot[0] = n;
}
void add(int x) {
  tot[cnt[x]]--;
  tot[++cnt[x]]++;
}
void del(int x) {
  tot[cnt[x]]--;
  tot[--cnt[x]]++;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), js[++cntjs] = a[i];
  for (int i = 1, tp, l, r; i <= q; i++) {
    scanf("%d%d%d", &tp, &l, &r);
    if (tp == 1)
      ask[++cntq] = (pos){cntq, l, r, cntu};
    else
      upd[++cntu] = (node){l, r}, js[++cntjs] = r;
  }
  init();
  int l = 1, r = 0, t = 0, now = 0;
  for (int i = 1; i <= cntq; i++) {
    int ql = ask[i].l, qr = ask[i].r, qt = ask[i].t;
    ;
    while (l > ql) add(a[--l]);
    while (r < qr) add(a[++r]);
    while (l < ql) del(a[l++]);
    while (r > qr) del(a[r--]);
    while (t < qt) {
      ++t;
      if (l <= upd[t].x && upd[t].x <= r) del(a[upd[t].x]), add(upd[t].val);
      swap(a[upd[t].x], upd[t].val);
    }
    while (t > qt) {
      if (l <= upd[t].x && upd[t].x <= r) del(a[upd[t].x]), add(upd[t].val);
      swap(a[upd[t].x], upd[t].val);
      --t;
    }
    for (int i = 1; i <= n; i++)
      if (tot[i] == 0) {
        now = i;
        break;
      }
    ans[ask[i].id] = now;
  }
  for (int i = 1; i <= cntq; i++) printf("%d\n", ans[i]);
  return 0;
}