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 = 200005;
struct node {
  int l, r, time, id;
} st1[maxn], st2[maxn];
int n, m, p, cnt1, cnt2;
int l = 1, r, t = 0, ans = 1;
int w[maxn], cnt[maxn], f[maxn], s[maxn];
vector<int> v;
inline bool cmp(const node &a, const node &b) {
  return a.l / p == b.l / p
             ? (a.r / p == b.r / p
                    ? (a.r / p) & 1 ? a.time > b.time : a.time < b.time
                : (a.l / p) & 1 ? a.r > b.r
                                : a.r < b.r)
             : a.l < b.l;
}
inline void add(const int &x) {
  f[cnt[w[x]]]--;
  f[++cnt[w[x]]]++;
}
inline void del(const int &x) {
  f[cnt[w[x]]]--;
  f[--cnt[w[x]]]++;
}
inline void change(int x) {
  if (l <= st2[x].l && st2[x].l <= r) {
    del(st2[x].l);
    swap(w[st2[x].l], st2[x].r);
    add(st2[x].l);
  } else {
    swap(w[st2[x].l], st2[x].r);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  p = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    v.push_back(w[i]);
  }
  for (int i = 1; i <= m; i++) {
    int op, l, r;
    cin >> op >> l >> r;
    if (op == 1) {
      st1[++cnt1].l = l;
      st1[cnt1].r = r;
      st1[cnt1].time = cnt2;
      st1[cnt1].id = cnt1;
    } else {
      st2[++cnt2].l = l;
      st2[cnt2].r = r;
      v.push_back(r);
    }
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 1; i <= n; i++) {
    w[i] = lower_bound(v.begin(), v.end(), w[i]) - v.begin() + 1;
  }
  for (int i = 1; i <= cnt2; i++) {
    st2[i].r = lower_bound(v.begin(), v.end(), st2[i].r) - v.begin() + 1;
  }
  f[0] = 1 << 30;
  sort(st1 + 1, st1 + cnt1 + 1, cmp);
  for (int i = 1; i <= cnt1; i++) {
    while (l > st1[i].l) {
      add(--l);
    }
    while (r < st1[i].r) {
      add(++r);
    }
    while (l < st1[i].l) {
      del(l++);
    }
    while (r > st1[i].r) {
      del(r--);
    }
    while (t < st1[i].time) {
      change(++t);
    }
    while (t > st1[i].time) {
      change(t--);
    }
    ans = 1;
    for (; f[ans]; ans++)
      ;
    s[st1[i].id] = ans;
  }
  for (int i = 1; i <= cnt1; i++) {
    cout << s[i] << "\n";
  }
}