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 + 5;
int n, m, k, x, y, block, a[N], f[N], ff[N], b[N], ans[N], g[N], tc, tq, tt,
    l = 1, r, t;
struct change {
  int p, s, la;
  change() {}
  change(int _p, int _s, int _la) { p = _p, s = _s, la = _la; }
} c[N];
struct query {
  int l, r, t, i;
  query() {}
  query(int _l, int _r, int _t, int _i) { l = _l, r = _r, t = _t, i = _i; }
} q[N];
bool cmp(query a, query b) {
  if (g[a.l] ^ g[b.l]) return g[a.l] < g[b.l];
  if (g[a.r] ^ g[b.r]) return g[a.r] < g[b.r];
  return a.t < b.t;
}
void del(int x) { f[x]--, ff[f[x] + 1]--, ff[f[x]]++; }
void add(int x) { f[x]++, ff[f[x] - 1]--, ff[f[x]]++; }
int calc() {
  for (int i = 1;; i++)
    if (!ff[i]) return i;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[++tt] = a[i];
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &k, &x, &y);
    if (k == 2)
      c[++tc] = change(x, y, a[x]), a[x] = y, b[++tt] = y;
    else
      q[++tq] = query(x, y, tc, tq);
  }
  sort(b + 1, b + 1 + tt);
  tt = unique(b + 1, b + 1 + tt) - b - 1;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + 1 + tt, a[i]) - b;
  for (int i = 1; i <= tc; i++)
    c[i].s = lower_bound(b + 1, b + 1 + tt, c[i].s) - b,
    c[i].la = lower_bound(b + 1, b + 1 + tt, c[i].la) - b;
  block = ceil(exp((log(n) + log(tc)) / 3));
  for (int i = 1; i <= n; i++) g[i] = (i - 1) / block + 1;
  for (int i = tc; i >= 1; i--) a[c[i].p] = c[i].la;
  sort(q + 1, q + 1 + tq, cmp);
  for (int i = 1; i <= tq; i++) {
    while (l < q[i].l) del(a[l++]);
    while (l > q[i].l) add(a[--l]);
    while (r < q[i].r) add(a[++r]);
    while (r > q[i].r) del(a[r--]);
    while (t < q[i].t) {
      int p = c[++t].p;
      if (l <= p && p <= r) del(a[p]);
      a[p] = c[t].s;
      if (l <= p && p <= r) add(a[p]);
    }
    while (t > q[i].t) {
      int p = c[t].p;
      if (l <= p && p <= r) del(a[p]);
      a[p] = c[t--].la;
      if (l <= p && p <= r) add(a[p]);
    }
    ans[q[i].i] = calc();
  }
  for (int i = 1; i <= tq; i++) printf("%d\n", ans[i]);
}