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 maxn = 2e5 + 10;
const double IO = 2.0 / 3.0;
int Ki[maxn] = {0}, l = 1, r = 0, m, n, Ans[maxn] = {0}, a[maxn] = {0},
    now[maxn] = {0}, num[maxn] = {0}, cnt[maxn] = {0}, tot = 0;
map<int, int> mp;
int f(int x) {
  if (!mp.count(x)) {
    mp[x] = ++tot;
  }
  return mp[x];
}
struct N1 {
  int l, r, id, tim;
} qu[maxn];
struct N2 {
  int pos, nxt, pre;
} ca[maxn];
bool cmp(N1 a, N1 b) {
  if (Ki[a.l] == Ki[b.l]) {
    if (Ki[a.r] == Ki[b.r]) return a.tim < b.tim;
    return a.r < b.r;
  }
  return a.l < b.l;
}
int tran() {
  for (int i = 1; 1; i++)
    if (cnt[i] == 0) {
      return i;
    }
}
void Add(int x, int d) {
  if (num[x] > 0) cnt[num[x]]--;
  num[x] += d;
  if (num[x] > 0) cnt[num[x]]++;
}
void Go(int x, int d) {
  if (l <= x && x <= r) {
    Add(a[x], -1);
    Add(d, 1);
  }
  a[x] = d;
}
int main(void) {
  mp.clear();
  scanf("%d%d", &n, &m);
  int unit = (int)pow(n, IO);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = now[i] = f(x);
    Ki[i] = i / unit + 1;
  }
  int t1 = 0, t2 = 0;
  for (int i = 1; i <= m; i++) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      qu[++t1] = N1{x, y, t1, t2};
    } else {
      y = f(y);
      ca[++t2] = N2{x, y, now[x]};
      now[x] = y;
    }
  }
  sort(qu + 1, qu + 1 + t1, cmp);
  int T = 0;
  for (int i = 1; i <= t1; i++) {
    while (T < qu[i].tim) {
      Go(ca[T + 1].pos, ca[T + 1].nxt);
      T++;
    }
    while (T > qu[i].tim) {
      Go(ca[T].pos, ca[T].pre);
      T--;
    }
    while (l < qu[i].l) {
      Add(a[l], -1);
      l++;
    }
    while (l > qu[i].l) {
      Add(a[l - 1], 1);
      l--;
    }
    while (r < qu[i].r) {
      Add(a[r + 1], 1);
      r++;
    }
    while (r > qu[i].r) {
      Add(a[r], -1);
      r--;
    }
    Ans[qu[i].id] = tran();
  }
  for (int i = 1; i <= t1; i++) printf("%d\n", Ans[i]);
  return 0;
}