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>
const int maxn = 2e5 + 10;
using namespace std;
int a[maxn], now[maxn], be[maxn], num[maxn], cnt[maxn], ans[maxn];
int n, m, l, r, t;
int msz;
map<int, int> M;
struct Query {
  int l, r, t, id;
} q[maxn];
int cmp(Query n1, Query n2) {
  if (be[n1.l] == be[n2.l]) {
    if (be[n1.r] == be[n2.r]) return n1.t < n2.t;
    return n1.r < n2.r;
  }
  return n1.l < n2.l;
}
struct nn {
  int pos, neww, oldd;
} c[maxn];
int compress(int x) {
  if (!M.count(x)) M[x] = ++msz;
  return M[x];
}
void add(int val, int d) {
  if (num[val] > 0) cnt[num[val]]--;
  num[val] += d;
  if (num[val] > 0) cnt[num[val]]++;
}
void go(int idx, int val) {
  if (l <= idx && idx <= r) {
    add(a[idx], -1);
    add(val, 1);
  }
  a[idx] = val;
}
int get() {
  for (int i = 1;; ++i) {
    if (cnt[i] == 0) return i;
  }
}
int main() {
  msz = 0;
  int x, qsz = 0, csz = 0, y, op;
  scanf("%d%d", &n, &m);
  int tt = pow(n, 2.0 / 3.0);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    now[i] = a[i] = compress(x);
    be[i] = (i - 1) / tt + 1;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1)
      q[++qsz] = (Query){x, y, csz, qsz};
    else {
      y = compress(y);
      c[++csz] = (nn){x, y, now[x]};
      now[x] = y;
    }
  }
  sort(q + 1, q + qsz + 1, cmp);
  l = 1, r = 0, t = 0;
  for (int i = 1; i <= qsz; i++) {
    while (t < q[i].t) go(c[t + 1].pos, c[t + 1].neww), ++t;
    while (t > q[i].t) go(c[t].pos, c[t].oldd), --t;
    while (l < q[i].l) add(a[l++], -1);
    while (l > q[i].l) add(a[--l], 1);
    while (r < q[i].r) add(a[++r], 1);
    while (r > q[i].r) add(a[r--], -1);
    ans[q[i].id] = get();
  }
  for (int i = 1; i <= qsz; i++) printf("%d\n", ans[i]);
}