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>
#pragma GCC optimize("inline")
int rd() {
  register int k = 0;
  char c = getchar();
  while (c > '9' || c < '0') c = getchar();
  while (c >= '0' && c <= '9') k = k * 10 + c - 48, c = getchar();
  return k;
}
const int N = 100001;
struct Q {
  int o, t, l, r;
} q[N];
struct C {
  int x, v;
} c[N];
int n, m, p, l = 1, r, qn, cn, a[N], bl[N], ans[N], T, b[N << 1], B, v[N << 1],
             sv[N], s;
bool cmp(register const Q& x, register const Q& y) {
  return bl[x.l] != bl[y.l] ? x.l < y.l
                            : (bl[x.r] == bl[y.r] ? x.t < y.t : x.r < y.r);
}
void Ins(register int p) { --sv[v[p]], ++sv[++v[p]]; }
void Del(register int p) { --sv[v[p]], ++sv[--v[p]]; }
void Set(register int t, register int i) {
  if (q[i].l <= c[t].x && c[t].x <= q[i].r) Del(a[c[t].x]), Ins(c[t].v);
  std::swap(a[c[t].x], c[t].v);
}
int main() {
  B = n = rd(), m = rd(), p = pow(n, 0.66);
  for (register int i = 1; i <= n; ++i) b[i] = a[i] = rd(), bl[i] = i / p;
  for (register int i = 1; i <= m; ++i) {
    if (rd() == 1)
      ++qn, q[qn] = {qn, cn, rd(), rd()};
    else
      c[++cn] = {rd(), b[++B] = rd()};
  }
  std::sort(b + 1, b + B + 1), B = std::unique(b + 1, b + B + 1) - b - 1,
                               std::sort(q + 1, q + qn + 1, cmp);
  for (register int i = 1; i <= n; ++i)
    a[i] = std::lower_bound(b + 1, b + B + 1, a[i]) - b;
  for (register int i = 1; i <= cn; ++i)
    c[i].v = std::lower_bound(b + 1, b + B + 1, c[i].v) - b;
  for (register int i = 1; i <= qn; ++i) {
    while (l > q[i].l) Ins(a[--l]);
    while (r < q[i].r) Ins(a[++r]);
    while (l < q[i].l) Del(a[l++]);
    while (r > q[i].r) Del(a[r--]);
    while (T < q[i].t) Set(++T, i);
    while (T > q[i].t) Set(T--, i);
    for (s = 1; sv[s] > 0; ++s)
      ;
    ans[q[i].o] = s;
  }
  for (register int i = 1; i <= qn; ++i) printf("%d\n", ans[i]);
  return 0;
}