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 N = 1e6 + 10;
map<int, int> key;
int n, m, cnt[N], Ans[N], a[N], Cnt[N];
int num, tot1, tot2;
struct node {
  int l, r, opl, opr, pre, id;
} q[N];
struct f {
  int pre, suf, id;
} cg[N];
bool cmp(register node x, register node y) {
  return x.opl != y.opl ? x.opl < y.opl
                        : (x.opr != y.opr ? x.opr < y.opr : x.pre < y.pre);
}
void inline add(int x) {
  Cnt[cnt[a[x]]]--;
  Cnt[++cnt[a[x]]]++;
}
void inline del(int x) {
  Cnt[cnt[a[x]]]--;
  Cnt[--cnt[a[x]]]++;
}
int main() {
  scanf("%d%d", &n, &m);
  int size = (int)pow(n, 2.0 / 3.0);
  for (register int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (!key[x]) key[x] = ++num;
    a[i] = key[x];
  }
  for (register int i = 1; i <= m; i++) {
    int x, l, r;
    scanf("%d%d%d", &x, &l, &r);
    if (x == 1) {
      q[++tot1].l = l;
      q[tot1].r = r;
      q[tot1].id = tot1;
      q[tot1].pre = tot2;
      q[tot1].opl = (q[tot1].l - 1) / size + 1;
      q[tot1].opr = (q[tot1].r - 1) / size + 1;
    } else {
      if (!key[r]) key[r] = ++num;
      r = key[r];
      cg[++tot2].id = l;
      cg[tot2].suf = r;
      cg[tot2].pre = a[cg[tot2].id];
      a[cg[tot2].id] = cg[tot2].suf;
    }
  }
  for (register int i = tot2; i >= 1; i--) a[cg[i].id] = cg[i].pre;
  sort(q + 1, q + tot1 + 1, cmp);
  int l = 1, r = 0, time = 0;
  for (register int i = 1; i <= tot1; i++) {
    while (l > q[i].l) add(--l);
    while (l < q[i].l) del(l++);
    while (r < q[i].r) add(++r);
    while (r > q[i].r) del(r--);
    while (q[i].pre < time) {
      int x = cg[time].id;
      if (l <= x && x <= r) del(x);
      a[x] = cg[time--].pre;
      if (l <= x && x <= r) add(x);
    }
    while (q[i].pre > time) {
      int x = cg[++time].id;
      if (l <= x && x <= r) del(x);
      a[x] = cg[time].suf;
      if (l <= x && x <= r) add(x);
    }
    int x = 1;
    while (Cnt[x]) x++;
    Ans[q[i].id] = x;
  }
  for (register int i = 1; i <= tot1; i++) printf("%d\n", Ans[i]);
  return 0;
}