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 = 200003;
int n, m, blo, b[maxn], tot, cntq, cntr, num[maxn], vis[maxn], cnt[maxn],
    a[maxn];
struct node {
  int l, r, id, c;
} q[maxn];
struct node2 {
  int p, c;
} r[maxn];
bool cmp(node a, node b) {
  return a.l / blo == b.l / blo
             ? (a.r / blo == b.r / blo ? a.c < b.c : a.r < b.r)
             : a.l < b.l;
}
void change(int x, int y) {
  if (q[y].l <= r[x].p && r[x].p <= q[y].r)
    cnt[vis[r[x].c]]--, cnt[++vis[r[x].c]]++, cnt[vis[a[r[x].p]]]--,
        cnt[--vis[a[r[x].p]]]++;
  swap(a[r[x].p], r[x].c);
}
void add(int x) { cnt[vis[a[x]]]--, cnt[++vis[a[x]]]++; }
void sub(int x) { cnt[vis[a[x]]]--, cnt[--vis[a[x]]]++; }
int main() {
  scanf("%d%d", &n, &m), blo = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), b[i] = a[i];
  for (int i = (tot = n, 1), opt, u, v; i <= m; ++i) {
    scanf("%d%d%d", &opt, &u, &v);
    if (opt == 1)
      q[++cntq] = (node){u, v, cntq, cntr};
    else
      r[++cntr] = (node2){u, v}, b[++tot] = v;
  }
  sort(b + 1, b + tot + 1);
  tot = unique(b + 1, b + tot + 1) - b - 1;
  for (int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + tot + 1, a[i]) - b;
  for (int i = 1; i <= cntr; ++i)
    r[i].c = lower_bound(b + 1, b + tot + 1, r[i].c) - b;
  sort(q + 1, q + cntq + 1, cmp);
  for (int i = 1, ql = 1, qr = 0, k = 0; i <= cntq; ++i) {
    int lef = q[i].l, rig = q[i].r;
    while (ql > lef) add(--ql);
    while (qr < rig) add(++qr);
    while (ql < lef) sub(ql++);
    while (qr > rig) sub(qr--);
    while (q[i].c > k) change(++k, i);
    while (q[i].c < k) change(k--, i);
    num[q[i].id] = 1;
    while (cnt[num[q[i].id]]) ++num[q[i].id];
  }
  for (int i = 1; i <= cntq; ++i) cout << num[i] << '\n';
  return 0;
}