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 long long inf = INT_MAX, df = 5e5 + 7;
long long i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z, buc[df],
    cnt[df], a[df], b[df], c[df], qcnt, tim, tot, ans[df];
struct qry {
  long long le, rig, l, id, tm;
} ask[df];
bool cmp(qry st, qry ch) {
  return (st.l == ch.l) ? (((st.rig / k == ch.rig / k) ? (st.tm < ch.tm)
                            : (st.l & 1)               ? st.rig > ch.rig
                                                       : st.rig < ch.rig))
                        : st.le < ch.le;
}
struct upd {
  long long p, x;
} cur[df];
inline long long read() {
  long long x = 0, y = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') y = (ch == '-') ? -1 : 1, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * y;
}
void add(long long x) {
  x = a[x];
  --cnt[buc[x]];
  ++buc[x];
  ++cnt[buc[x]];
}
void det(long long x) {
  x = a[x];
  --cnt[buc[x]];
  --buc[x];
  ++cnt[buc[x]];
}
void modity(long long t, long long l, long long r) {
  if (cur[t].p >= l && cur[t].p <= r) {
    det(cur[t].p);
    swap(a[cur[t].p], cur[t].x);
    add(cur[t].p);
    return;
  }
  swap(a[cur[t].p], cur[t].x);
  return;
}
int main() {
  n = read(), q = read();
  k = 3000;
  for (long long i = (1); i <= (n); ++i) b[i] = a[i] = read();
  sort(b + 1, b + n + 1), tot = unique(b + 1, b + n + 1) - b - 1;
  for (long long i = (1); i <= (q); ++i) {
    long long op = read();
    if (op == 1) {
      ask[++qcnt].le = read(), ask[qcnt].rig = read(), ask[qcnt].id = qcnt,
      ask[qcnt].tm = tim;
      ask[qcnt].l = ask[qcnt].le / k;
    }
    if (op == 2) {
      cur[++tim].p = read(), cur[tim].x = read();
      b[++tot] = cur[tim].x;
    }
  }
  sort(b + 1, b + tot + 1);
  tot = unique(b + 1, b + tot + 1) - b - 1;
  for (long long i = (1); i <= (n); ++i)
    a[i] = lower_bound(b + 1, b + tot + 1, a[i]) - b;
  for (long long i = (1); i <= (tim); ++i) {
    cur[i].x = lower_bound(b + 1, b + tot + 1, cur[i].x) - b;
  }
  sort(ask + 1, ask + qcnt + 1, cmp);
  long long l = 1, r = 0, tim = 0;
  for (long long i = (1); i <= (qcnt); ++i) {
    while (r < ask[i].rig) add(++r);
    while (r > ask[i].rig) det(r--);
    while (l < ask[i].le) det(l++);
    while (l > ask[i].le) add(--l);
    while (tim < ask[i].tm) modity(++tim, l, r);
    while (tim > ask[i].tm) modity(tim--, l, r);
    for (ans[ask[i].id] = 1; cnt[ans[ask[i].id]]; ++ans[ask[i].id])
      ;
  }
  for (long long i = (1); i <= (qcnt); ++i) printf("%lld\n", ans[i]);
}