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;
inline int read() {
  int x = 0;
  bool f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = 0;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  if (f) return x;
  return 0 - x;
}
const int maxn = 200000 + 10;
int n, m, col[maxn], block, ans[maxn], cnt[maxn], last[maxn], sum[maxn];
struct node {
  int l, r, id, t;
  bool operator<(const node &x) const {
    if (((l - 1) / block + 1) == ((x.l - 1) / block + 1)) {
      if (((r - 1) / block + 1) == ((x.r - 1) / block + 1)) return t < x.t;
      return r < x.r;
    }
    return l < x.l;
  }
} q[maxn];
struct modify {
  int x, old, nxt;
} qq[maxn];
int c1 = 0, c2 = 0, p = 0;
map<int, int> mp;
int get(int x) {
  if (mp[x] == 0) mp[x] = ++p;
  return mp[x];
}
void add(int x) {
  sum[cnt[x]]--;
  cnt[x]++;
  sum[cnt[x]]++;
}
void del(int x) {
  sum[cnt[x]]--;
  cnt[x]--;
  sum[cnt[x]]++;
}
int l = 1, r = 0, t = 0;
void update(int x, int val) {
  if (l <= x && x <= r) {
    del(col[x]), add(val);
  }
  col[x] = val;
}
void solve() {
  sort(q + 1, q + c1 + 1);
  for (int i = 1; i <= c1; i++) {
    while (t < q[i].t) t++, update(qq[t].x, qq[t].nxt);
    while (t > q[i].t) update(qq[t].x, qq[t].old), t--;
    while (l < q[i].l) del(col[l]), l++;
    while (l > q[i].l) l--, add(col[l]);
    while (r < q[i].r) r++, add(col[r]);
    while (r > q[i].r) del(col[r]), r--;
    int ret = 1;
    while (sum[ret] != 0) ret++;
    ans[q[i].id] = ret;
  }
  for (int i = 1; i <= c1; i++) printf("%d\n", ans[i]);
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    col[i] = read();
    col[i] = last[i] = get(col[i]);
  }
  block = pow(n, 2.0 / 3);
  for (int i = 1, op, x, y; i <= m; i++) {
    op = read(), x = read(), y = read();
    if (op == 1) {
      q[++c1] = node{x, y, c1, c2};
    } else {
      y = get(y);
      qq[++c2] = modify{x, last[x], y};
      last[x] = y;
    }
  }
  solve();
  return 0;
}