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;
struct QNode {
  int l, r;
  int sn;
  int t;
};
struct CNode {
  int p, v;
};
int n, m;
int a[200010];
int b[200010];
int block;
QNode q[200010];
CNode c[200010];
int qn, cn;
map<int, int> mp;
int cnt[200010];
int num[200010];
int ans[200010];
bool cmp(QNode q1, QNode q2) {
  if (b[q1.l] == b[q2.l]) {
    if (b[q1.r] == b[q2.r]) {
      return q1.t < q2.t;
    }
    return q1.r < q2.r;
  }
  return q1.l < q2.l;
}
void add(int x) {
  num[cnt[x]]--;
  cnt[x]++;
  num[cnt[x]]++;
}
void del(int x) {
  num[cnt[x]]--;
  cnt[x]--;
  num[cnt[x]]++;
}
void change(int t, int i) {
  if (c[t].p >= q[i].l && c[t].p <= q[i].r) {
    del(a[c[t].p]);
    add(c[t].v);
  }
  swap(c[t].v, a[c[t].p]);
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  block = pow(n, 2.0 / 3);
  int w = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (mp[a[i]] == 0) {
      w++;
      mp[a[i]] = w;
    }
    a[i] = mp[a[i]];
    b[i] = (i - 1) / block + 1;
  }
  for (int i = 1; i <= m; i++) {
    int k;
    cin >> k;
    if (k == 1) {
      int l, r;
      cin >> l >> r;
      qn++;
      q[qn].l = l;
      q[qn].r = r;
      q[qn].sn = qn;
      q[qn].t = cn;
    } else {
      int p, v;
      cin >> p >> v;
      if (mp[v] == 0) {
        w++;
        mp[v] = w;
      }
      cn++;
      c[cn].p = p;
      c[cn].v = mp[v];
    }
  }
  sort(q + 1, q + qn + 1, cmp);
  int l = 1, r = 0, t = 0;
  for (int i = 1; i <= qn; i++) {
    while (r < q[i].r) {
      r++;
      add(a[r]);
    }
    while (r > q[i].r) {
      del(a[r]);
      r--;
    }
    while (l > q[i].l) {
      l--;
      add(a[l]);
    }
    while (l < q[i].l) {
      del(a[l]);
      l++;
    }
    while (t < q[i].t) {
      t++;
      change(t, i);
    }
    while (t > q[i].t) {
      change(t, i);
      t--;
    }
    for (int j = 1; j <= 100010; j++) {
      if (num[j] == 0) {
        ans[q[i].sn] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= qn; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}