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;
long long n, m;
long long siz, a[1000005];
long long ans[1000005];
long long cnt, cn[1000005];
struct node {
  long long l, r, pos, id;
  long long t;
  node() {}
  node(long long l, long long r, long long pos, long long id)
      : l(l), r(r), pos(pos), id(id) {}
};
node f[1000005], g[1000005];
bool com(node a, node b) {
  if (a.l / siz != b.l / siz) return a.l < b.l;
  if (a.r / siz != b.r / siz) return a.r < b.r;
  return a.t < b.t;
}
long long num[1000005];
void add(long long x) {
  --num[cn[x]];
  ++num[++cn[x]];
}
void del(long long x) {
  --num[cn[x]];
  ++num[--cn[x]];
}
void upd(long long x, long long t) {
  if (f[x].l <= g[t].l && g[t].l <= f[x].r) {
    del(a[g[t].l]);
    add(g[t].r);
  }
  swap(a[g[t].l], g[t].r);
}
vector<long long> vec;
int main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    vec.push_back(a[i]);
  }
  siz = pow(n, 0.66);
  long long s;
  long long k = 0;
  long long u = 0;
  long long p, q;
  for (long long i = 1; i <= m; i++) {
    scanf("%lld%lld%lld", &s, &p, &q);
    if (s == 1)
      f[++u].l = p, f[u].r = q, f[u].id = u, f[u].t = k;
    else {
      g[++k].l = p;
      g[k].r = q;
      vec.push_back(q);
    }
  }
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  for (long long i = 1; i <= n; i++) {
    a[i] = lower_bound(vec.begin(), vec.end(), a[i]) - vec.begin() + 1;
  }
  for (int i = 1; i <= k; i++)
    g[i].r = lower_bound(vec.begin(), vec.end(), g[i].r) - vec.begin() + 1;
  sort(f + 1, f + 1 + u, com);
  long long l = 1, r = 0, t = 0;
  num[0] = 1e9;
  for (long long i = 1; i <= u; i++) {
    while (l > f[i].l) add(a[--l]);
    while (r < f[i].r) add(a[++r]);
    while (l < f[i].l) del(a[l++]);
    while (r > f[i].r) del(a[r--]);
    while (t < f[i].t) upd(i, ++t);
    while (t > f[i].t) upd(i, t--);
    for (long long j = 1; j <= n + 10; j++) {
      if (num[j] == 0) {
        ans[f[i].id] = j;
        break;
      }
    }
  }
  for (long long i = 1; i <= u; i++) printf("%lld\n", ans[i]);
  return 0;
}