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 que {
  int l, r, t, id, ans;
} q[100005];
struct modify {
  int pos, r, x;
} mo[100005];
int n, m, m1, m2;
int a[100005], tt[100005];
int tmp[200005], vis[200005], tp;
int vi2[200005];
int dk[200005], K;
int cmp(que x, que y) {
  if (dk[x.l] != dk[y.l]) return dk[x.l] < dk[y.l];
  if (dk[x.r] != dk[y.r]) return dk[x.r] < dk[y.r];
  return x.t < y.t;
}
int cmp2(que x, que y) { return x.id < y.id; }
void add(int p) {
  vi2[vis[a[p]]]--;
  vis[a[p]]++;
  vi2[vis[a[p]]]++;
}
void del(int p) {
  vi2[vis[a[p]]]--;
  vis[a[p]]--;
  vi2[vis[a[p]]]++;
}
void solve() {
  int l = 1, r = 0, t = 0;
  for (int i = 1; i <= m1; i++) {
    while (l > q[i].l) {
      add(--l);
    }
    while (r < q[i].r) {
      add(++r);
    }
    while (l < q[i].l) {
      del(l++);
    }
    while (r > q[i].r) {
      del(r--);
    }
    while (t < q[i].t) {
      t++;
      if (mo[t].pos >= l && mo[t].pos <= r) {
        del(mo[t].pos);
      }
      a[mo[t].pos] = mo[t].x;
      if (mo[t].pos >= l && mo[t].pos <= r) {
        add(mo[t].pos);
      }
    }
    while (t > q[i].t) {
      if (mo[t].pos >= l && mo[t].pos <= r) {
        del(mo[t].pos);
      }
      a[mo[t].pos] = mo[t].r;
      if (mo[t].pos >= l && mo[t].pos <= r) {
        add(mo[t].pos);
      }
      t--;
    }
    q[i].ans = 1;
    while (vi2[q[i].ans]) q[i].ans++;
  }
}
map<int, int> mp;
int newid;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (!mp[a[i]]) mp[a[i]] = ++newid;
    a[i] = mp[a[i]];
    tt[i] = a[i];
  }
  for (int i = 1; i <= m; i++) {
    int opt, l, r;
    scanf("%d%d%d", &opt, &l, &r);
    if (opt == 1) {
      m1++;
      q[m1].l = l;
      q[m1].r = r;
      q[m1].t = m2;
      q[m1].id = i;
    }
    if (opt == 2) {
      m2++;
      mo[m2].pos = l;
      if (!mp[r]) mp[r] = ++newid;
      mo[m2].x = mp[r];
      mo[m2].r = tt[l];
      tt[l] = mp[r];
    }
  }
  K = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) {
    dk[i] = (i - 1) / K + 1;
  }
  sort(q + 1, q + m1 + 1, cmp);
  solve();
  sort(q + 1, q + m1 + 1, cmp2);
  for (int i = 1; i <= m1; i++) {
    printf("%d\n", q[i].ans);
  }
  return 0;
}