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 len = 2e5 + 10;
const double pi = acos(-1.0);
const int mod = 1e4 + 7;
const int inf = 0x3f3f3f3f;
struct Q {
  int l, r, id;
  int pre;
} q[len];
int qn;
struct C {
  int pos, v;
} c[len];
int cn;
int n, m, block;
int a[len], L[len], R[len];
vector<int> ve;
int get(int x) { return lower_bound(ve.begin(), ve.end(), x) - ve.begin() + 1; }
bool cmp(Q a, Q b) {
  if ((a.l - 1) / block != (b.l - 1) / block)
    return (a.l - 1) / block < (b.l - 1) / block;
  if ((a.r - 1) / block != (b.r - 1) / block)
    return (a.r - 1) / block < (b.r - 1) / block;
  return a.pre < b.pre;
}
int num1[len], num2[len], out[len];
void update(int x, int v) {
  int h = a[x];
  num2[num1[h]]--;
  num1[h] += v;
  num2[num1[h]]++;
}
void change(int now, int i) {
  if (q[i].l <= c[now].pos && c[now].pos <= q[i].r) {
    int h = a[c[now].pos];
    num2[num1[h]]--;
    num1[h]--;
    num2[num1[h]]++;
    h = c[now].v;
    num2[num1[h]]--;
    num1[h]++;
    num2[num1[h]]++;
  }
  swap(a[c[now].pos], c[now].v);
}
void solve() {
  int l = 1, r = 0, now = 0;
  for (int i = 1; i <= qn; ++i) {
    while (l > q[i].l) update(l - 1, 1), l--;
    while (l < q[i].l) update(l, -1), l++;
    while (r > q[i].r) update(r, -1), r--;
    while (r < q[i].r) update(r + 1, 1), r++;
    while (now < q[i].pre) change(++now, i);
    while (now > q[i].pre) change(now--, i);
    for (int j = 1;; j++)
      if (num2[j] == 0) {
        out[q[i].id] = j;
        break;
      }
  }
  for (int i = 1; i <= qn; ++i) printf("%d\n", out[i]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    ve.push_back(a[i]);
  }
  block = pow(1.0 * n, 2.0 / 3.0);
  for (int i = 1; i <= m; ++i) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      ++qn;
      scanf("%d%d", &q[qn].l, &q[qn].r);
      q[qn].pre = cn;
      q[qn].id = qn;
    } else if (op == 2) {
      ++cn;
      scanf("%d%d", &c[cn].pos, &c[cn].v);
      ve.push_back(c[cn].v);
    }
  }
  sort(ve.begin(), ve.end());
  ve.erase(unique(ve.begin(), ve.end()), ve.end());
  for (int i = 1; i <= n; ++i) a[i] = get(a[i]);
  for (int i = 1; i <= cn; ++i) c[i].v = get(c[i].v);
  sort(q + 1, q + 1 + qn, cmp);
  solve();
}