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 = 1e5 + 1000;
int n, q, s, si[maxn], tq, tc, cur, temp, ctt[maxn * 2], li[maxn * 2],
    vis[maxn * 2], curl = 1, curr = 0, now, ans[maxn];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
typedef struct a {
  int l, r, id, pre;
} per;
per ask[maxn];
bool cmp(per a, per b) {
  return a.l / s == b.l / s
             ? a.r / s == b.r / s ? a.pre < b.pre : a.r / s < b.r / s
             : a.l / s < b.l / s;
}
typedef struct b {
  int pos, cou;
} node;
node change[maxn];
void add(int num) {
  vis[ctt[num]]--;
  ctt[num]++;
  vis[ctt[num]]++;
}
void del(int num) {
  if (num == 0) return;
  vis[ctt[num]]--;
  ctt[num]--;
  if (ctt[num] >= 0) vis[ctt[num]]++;
}
void work(int now, int i) {
  if (change[now].pos >= ask[i].l && change[now].pos <= ask[i].r) {
    add(change[now].cou);
    del(si[change[now].pos]);
  }
  swap(si[change[now].pos], change[now].cou);
}
int main() {
  n = read();
  q = read();
  s = pow(n, 0.666);
  for (int i = 1; i <= n; ++i) {
    si[i] = read();
    li[++cur] = si[i];
  }
  int sign;
  for (int i = 1; i <= q; ++i) {
    sign = read();
    if (sign == 1) {
      tq++;
      ask[tq].l = read();
      ask[tq].r = read();
      ask[tq].pre = tc;
      ask[tq].id = tq;
    } else {
      tc++;
      change[tc].pos = read();
      change[tc].cou = read();
      li[++cur] = change[tc].cou;
    }
  }
  sort(ask + 1, ask + 1 + tq, cmp);
  sort(li + 1, li + 1 + cur);
  int len = unique(li + 1, li + 1 + cur) - li - 1;
  for (int i = 1; i <= n; ++i)
    si[i] = lower_bound(li + 1, li + 1 + len, si[i]) - li;
  for (int i = 1; i <= tc; ++i)
    change[i].cou = lower_bound(li + 1, li + 1 + len, change[i].cou) - li;
  temp = 1;
  for (int i = 1; i <= tq; ++i) {
    int l = ask[i].l, r = ask[i].r, time = ask[i].pre;
    while (curr < r) add(si[++curr]);
    while (curl > l) add(si[--curl]);
    while (curr > r) del(si[curr--]);
    while (curl < l) del(si[curl++]);
    while (now < time) work(++now, i);
    while (now > time) work(now--, i);
    int tans = 1;
    while (vis[tans]) tans++;
    ans[ask[i].id] = tans;
  }
  for (int i = 1; i <= tq; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}