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;
typedef struct {
  int id;
  int l;
  int r;
  int t;
} Query;
typedef struct {
  int pos;
  int val;
} Replacement;
int block;
int a[100007], b[200007], cnt[200007], vis[200007], ans[100007];
Query query[100007];
Replacement replacement[100007];
bool operator<(const Query a, const Query b) {
  if (a.l / block != b.l / block) return a.l < b.l;
  if (a.r / block != b.r / block) return a.r < b.r;
  return a.t < b.t;
}
inline void del(int x) {
  cnt[vis[x]]--;
  cnt[--vis[x]]++;
}
inline void add(int x) {
  cnt[vis[x]]--;
  cnt[++vis[x]]++;
}
inline void release(int x, int t) {
  if (query[x].l <= replacement[t].pos && replacement[t].pos <= query[x].r) {
    del(a[replacement[t].pos]);
    add(replacement[t].val);
  }
  swap(a[replacement[t].pos], replacement[t].val);
}
int main() {
  int n, q, m, k = 0, p = 0;
  cin >> n >> q;
  m = n;
  block = n / cbrt(q);
  for (register int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  for (register int i = 1; i <= q; i++) {
    int op;
    cin >> op;
    if (op == 1) {
      k++;
      cin >> query[k].l >> query[k].r;
      query[k].id = k;
      query[k].t = p;
    } else {
      p++;
      cin >> replacement[p].pos >> replacement[p].val;
      b[++m] = replacement[p].val;
    }
  }
  sort(b + 1, b + m + 1);
  m = unique(b + 1, b + m + 1) - b - 1;
  for (register int i = 1; i <= n; i++) {
    a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
  }
  for (register int i = 1; i <= p; i++) {
    replacement[i].val = lower_bound(b + 1, b + m + 1, replacement[i].val) - b;
  }
  sort(query + 1, query + k + 1);
  for (register int i = 1, j = 1, x = 0, y = 0; i <= k; i++) {
    while (j < query[i].l) del(a[j++]);
    while (j > query[i].l) add(a[--j]);
    while (x < query[i].r) add(a[++x]);
    while (x > query[i].r) del(a[x--]);
    while (y < query[i].t) release(i, ++y);
    while (y > query[i].t) release(i, y--);
    ans[query[i].id] = 1;
    while (cnt[ans[query[i].id]] > 0) ans[query[i].id]++;
  }
  for (register int i = 1; i <= k; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}