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 = 2E5 + 7;
map<int, int> mp;
int cnt[N], vis[N], a[N];
int BLOCK;
int ans[N];
int num[N], tot;
int n, q;
int cntq;
int cntc;
struct QQ {
  int l, r, z, id;
} Q[N];
struct CC {
  int x, y;
} C[N];
bool cmp(QQ a, QQ b) {
  if (a.l / BLOCK == b.l / BLOCK) {
    if (a.r / BLOCK == b.r / BLOCK) return a.z < b.z;
    return a.r < b.r;
  } else {
    return a.l < b.l;
  }
}
void add(int x) {
  --vis[cnt[num[x]]];
  ++cnt[num[x]];
  ++vis[cnt[num[x]]];
}
void dec(int x) {
  --vis[cnt[num[x]]];
  --cnt[num[x]];
  ++vis[cnt[num[x]]];
}
void solve() {
  int l = 1, r = 1, c = 0, res;
  add(1);
  for (int i = 1; i <= cntq; ++i) {
    while (r < Q[i].r) add(++r);
    while (r > Q[i].r) dec(r--);
    while (l < Q[i].l) dec(l++);
    while (l > Q[i].l) add(--l);
    while (c > Q[i].z) {
      if (C[c].x >= l && C[c].x <= r) dec(C[c].x);
      swap(C[c].y, num[C[c].x]);
      if (C[c].x >= l && C[c].x <= r) add(C[c].x);
      c--;
    }
    while (c < Q[i].z) {
      c++;
      if (C[c].x >= l && C[c].x <= r) dec(C[c].x);
      swap(C[c].y, num[C[c].x]);
      if (C[c].x >= l && C[c].x <= r) add(C[c].x);
    }
    res = 1;
    while (vis[res]) ++res;
    ans[Q[i].id] = res;
  }
  for (int i = 1; i <= cntq; i++) printf("%d\n", ans[i]);
}
int main() {
  scanf("%d %d", &n, &q);
  BLOCK = (int)pow(n, 2.0 / 3.0);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    if (mp[x])
      num[i] = mp[x];
    else
      num[i] = mp[x] = ++tot;
  }
  int opt, l, r;
  for (int i = 1; i <= q; ++i) {
    scanf("%d %d %d", &opt, &l, &r);
    if (opt == 1) {
      ++cntq;
      Q[cntq] = (QQ){l, r, cntc, cntq};
    } else {
      ++cntc;
      if (mp[r])
        r = mp[r];
      else
        r = mp[r] = ++tot;
      C[cntc] = (CC){l, r};
    }
  }
  sort(Q + 1, Q + 1 + cntq, cmp);
  solve();
  return 0;
}