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;
int n, m, a[100010], mq, mc, opt, num[100010], L = 1, R, sz = 2154, chg;
int cnt[200010], maxn, st[200010], nx, rp;
struct que {
  int l, r, id, ch;
  bool operator<(const que v) const {
    if (l / sz != v.l / sz) return l < v.l;
    if (r / sz != v.r / sz) return r < v.r;
    return ch < v.ch;
  }
} q[100010];
struct ch {
  int x, y;
} c[100010];
struct numx {
  int x, id;
  bool operator<(const numx v) const { return x < v.x; }
} b[200010];
void add(int i) { --cnt[st[i]++], cnt[st[i]]++; }
void del(int i) { --cnt[st[i]--], ++cnt[st[i]]; }
void change(int i) {
  if (L <= c[i].x && c[i].x <= R) {
    del(a[c[i].x]);
    add(c[i].y);
  }
  swap(c[i].y, a[c[i].x]);
}
int solve() {
  for (int i = 1;; ++i)
    if (!cnt[i]) return i;
}
void read(int& u) {
  char x;
  while ((x = getchar()) < '0' || x > '9')
    ;
  u = x - '0';
  while ((x = getchar()) >= '0' && x <= '9') u = (u << 3) + (u << 1) + x - '0';
}
int main() {
  read(n), read(m);
  for (int i = 1; i <= n; ++i) read(b[i].x), b[i].id = i;
  nx = n;
  for (int i = 1; i <= m; ++i) {
    read(opt);
    if (opt == 1) {
      ++mq;
      read(q[mq].l), read(q[mq].r);
      q[mq].id = mq;
      q[mq].ch = mc;
    } else {
      ++mc;
      read(c[mc].x), read(c[mc].y);
      b[++nx].x = c[mc].y;
      b[nx].id = -mc;
    }
  }
  sort(b + 1, b + nx + 1);
  b[0].x = 0;
  for (int i = 1; i <= nx; ++i) {
    if (b[i].x != b[i - 1].x) maxn++;
    if (b[i].id > 0)
      a[b[i].id] = maxn;
    else
      c[-b[i].id].y = maxn;
  }
  sort(q + 1, q + mq + 1);
  cnt[0] = n << 1;
  for (int i = 1; i <= mq; ++i) {
    while (q[i].l < L) add(a[--L]);
    while (R < q[i].r) add(a[++R]);
    while (L < q[i].l) del(a[L++]);
    while (q[i].r < R) del(a[R--]);
    while (chg < q[i].ch) change(++chg);
    while (chg > q[i].ch) change(chg--);
    num[q[i].id] = solve();
  }
  for (int i = 1; i <= mq; ++i) printf("%d\n", num[i]);
  return 0;
}