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;
int N, M, LEN;
int A[200005], inb[200005];
int A1[200005], n;
int Pos[200005], Kbef[200005], Kaft[200005], cntc;
struct node {
  int l, r, id, c;
  bool operator<(const node b) const {
    return inb[l] < inb[b.l] || (inb[l] == inb[b.l] && inb[r] < inb[b.r]) ||
           (inb[l] == inb[b.l] && inb[r] == inb[b.r] && c < b.c);
  }
} Q[200005];
int cntq;
int Ans[200005];
int cnt1[200005];
int cnt2[200005];
int ANS;
void inc(int x) {
  cnt2[cnt1[x]]--;
  cnt1[x]++;
  cnt2[cnt1[x]]++;
}
void dec(int x) {
  cnt2[cnt1[x]]--;
  cnt1[x]--;
  cnt2[cnt1[x]]++;
}
int main() {
  scanf("%d%d", &N, &M);
  LEN = pow(N, 0.66667);
  for (int i = 1; i <= N; i++)
    scanf("%d", &A[i]), A1[++n] = A[i], inb[i] = (i - 1) / LEN + 1;
  for (int i = 1; i <= M; i++) {
    int opt;
    scanf("%d", &opt);
    if (opt == 1) {
      cntq++;
      Q[cntq].id = cntq;
      Q[cntq].c = cntc;
      scanf("%d%d", &Q[cntq].l, &Q[cntq].r);
    } else {
      cntc++;
      scanf("%d%d", &Pos[cntc], &Kaft[cntc]);
      Kbef[cntc] = A[Pos[cntc]];
      A[Pos[cntc]] = Kaft[cntc];
      A1[++n] = Kaft[cntc];
    }
  }
  for (int i = cntc; i >= 1; i--) A[Pos[i]] = Kbef[i];
  sort(A1 + 1, A1 + n + 1);
  n = unique(A1 + 1, A1 + n + 1) - A1 - 1;
  for (int i = 1; i <= N; i++)
    A[i] = lower_bound(A1 + 1, A1 + n + 1, A[i]) - A1;
  for (int i = 1; i <= cntc; i++)
    Kbef[i] = lower_bound(A1 + 1, A1 + n + 1, Kbef[i]) - A1,
    Kaft[i] = lower_bound(A1 + 1, A1 + n + 1, Kaft[i]) - A1;
  sort(Q + 1, Q + cntq + 1);
  int L = 1, R = 1, C = 0;
  inc(A[1]);
  cnt2[0] = 400005;
  for (int i = 1; i <= cntq; i++) {
    while (R < Q[i].r) inc(A[++R]);
    while (L > Q[i].l) inc(A[--L]);
    while (R > Q[i].r) dec(A[R--]);
    while (L < Q[i].l) dec(A[L++]);
    while (C < Q[i].c) {
      C++;
      if (Pos[C] >= L && Pos[C] <= R) inc(Kaft[C]), dec(Kbef[C]);
      A[Pos[C]] = Kaft[C];
    }
    while (C > Q[i].c) {
      if (Pos[C] >= L && Pos[C] <= R) inc(Kbef[C]), dec(Kaft[C]);
      A[Pos[C]] = Kbef[C];
      C--;
    }
    ANS = 0;
    while (cnt2[ANS]) ANS++;
    Ans[Q[i].id] = ANS;
  }
  for (int i = 1; i <= cntq; i++) printf("%d\n", Ans[i]);
  return 0;
}