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 + 10;
int a[maxn], num[maxn << 1], c[maxn], last[maxn];
int cnt1, cnt2, cnt;
int block;
int L = 1, R = 0;
int A[maxn];
map<int, int> vis;
int get(int x) { return (x - 1) / block + 1; }
struct node {
  int l, r, tim, id;
  int lb, rb;
  bool operator<(const node& x) const {
    if (lb != x.lb) return l < x.l;
    if (rb != x.rb) return r < x.r;
    return tim < x.tim;
  }
} s[maxn];
struct Node {
  int pos, col, pre;
} ss[maxn];
void add(int val) {
  c[num[val]]--;
  num[val]++;
  c[num[val]]++;
}
void del(int val) {
  c[num[val]]--;
  num[val]--;
  c[num[val]]++;
}
void change(int pos, int col) {
  if (L <= pos && R >= pos) {
    del(a[pos]);
    add(col);
  }
  a[pos] = col;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  block = 2100;
  for (int i = 1; i <= n; i++) {
    int val;
    scanf("%d", &val);
    if (!vis[val]) vis[val] = ++cnt;
    val = vis[val];
    a[i] = val;
    last[i] = a[i];
  }
  for (int i = 1; i <= m; i++) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      s[++cnt1].l = x;
      s[cnt1].r = y;
      s[cnt1].id = cnt1;
      s[cnt1].tim = cnt2;
      s[cnt1].lb = get(x);
      s[cnt1].rb = get(y);
    } else {
      if (!vis[y]) vis[y] = ++cnt;
      ss[++cnt2].pos = x;
      ss[cnt2].col = vis[y];
      ss[cnt2].pre = last[x];
      last[x] = vis[y];
    }
  }
  sort(s + 1, s + 1 + cnt1);
  int pl = 0;
  for (int i = 1; i <= cnt1; i++) {
    int l = s[i].l, r = s[i].r;
    int tim = s[i].tim, id = s[i].id;
    while (pl < tim) {
      pl++;
      change(ss[pl].pos, ss[pl].col);
    }
    while (pl > tim) {
      change(ss[pl].pos, ss[pl].pre);
      pl--;
    }
    while (L > l) add(a[--L]);
    while (R < r) add(a[++R]);
    while (L < l) del(a[L++]);
    while (R > r) del(a[R--]);
    for (int j = 1;; j++)
      if (!c[j]) {
        A[id] = j;
        break;
      }
  }
  for (int i = 1; i <= cnt1; i++) printf("%d\n", A[i]);
  return 0;
}