940_F. Machine Learning

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.

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


Example output:
2
3
2


Tags: brute force
data structures
Difficulty rating: 2600

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, q, k, l, r, t, m1, m2, op, tot, v[100003], s[100003], ans[100003],
    cnt[100003 * 2];
map<int, int> u;
struct query {
  int l, r, t, lk, rk, id;
} a[100003];
struct modify {
  int pos, v0, v1;
} b[100003];
int uni(int x) { return u.count(x) ? u[x] : u[x] = ++tot; }
bool cmp(query a, query b) {
  if (a.lk != b.lk) return a.lk < b.lk;
  if (a.rk != b.rk) return a.rk < b.rk;
  return a.t < b.t;
}
void add(int x) {
  s[cnt[x]]--;
  s[++cnt[x]]++;
}
void del(int x) {
  s[cnt[x]]--;
  s[--cnt[x]]++;
}
void mo(int pos, int x) {
  if (pos >= l && pos <= r) {
    add(x);
    del(v[pos]);
  }
  v[pos] = x;
}
int get(int nl, int nr, int nt) {
  while (t < nt) {
    ++t;
    mo(b[t].pos, b[t].v1);
  }
  while (t > nt) {
    mo(b[t].pos, b[t].v0);
    --t;
  }
  while (l > nl) add(v[--l]);
  while (r < nr) add(v[++r]);
  while (l < nl) del(v[l++]);
  while (r > nr) del(v[r--]);
  int i;
  for (i = 1; s[i]; i++)
    ;
  return i;
}
int main() {
  scanf("%d%d", &n, &q);
  k = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    v[i] = uni(v[i]);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1)
      a[++m1] = (query){l, r, m2, (l - 1) / k, (r - 1) / k, m1};
    else
      b[++m2].pos = l, b[m2].v1 = uni(r);
  }
  for (int i = 1; i <= m2; i++) {
    b[i].v0 = v[b[i].pos];
    v[b[i].pos] = b[i].v1;
  }
  sort(a + 1, a + m1 + 1, cmp);
  t = m2, l = 1, r = 0;
  for (int i = 1; i <= m1; i++) ans[a[i].id] = get(a[i].l, a[i].r, a[i].t);
  for (int i = 1; i <= m1; i++) printf("%d\n", ans[i]);
  return 0;
}