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 bl = 2333;
struct query {
  int ans, l, r, t, lbl, rbl, id;
} b[200005];
pair<int, int> c[200005];
int now = 0;
int cnt[200005], vis[200005], a[200005], s[200005], Q, pos[200005];
inline bool cmp1(query x, query y) {
  return x.lbl < y.lbl || x.lbl == y.lbl && x.rbl < y.rbl ||
         x.lbl == y.lbl && x.rbl == y.rbl && x.t < y.t;
}
inline bool cmp2(query x, query y) { return x.id < y.id; }
inline void add(int x) { --vis[cnt[x]], ++cnt[x], ++vis[cnt[x]]; }
inline void del(int x) { --vis[cnt[x]], --cnt[x], ++vis[cnt[x]]; }
inline void change(int x, int l, int r) {
  if (!c[x].first) return;
  if (pos[x] <= r && pos[x] >= l) del(a[pos[x]]), add(c[x].second);
  a[pos[x]] = c[x].second;
}
inline void undo(int x, int l, int r) {
  if (!c[x].first) return;
  if (pos[x] <= r && pos[x] >= l) del(a[pos[x]]), add(c[x].first);
  a[pos[x]] = c[x].first;
}
map<int, int> mp;
int main(int argc, char** argv) {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!mp[a[i]]) mp[a[i]] = ++now;
    a[i] = mp[a[i]];
    s[i] = a[i];
  }
  for (int i = 1; i <= m; i++) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      b[++Q] = {1, l, r, i, l / bl, r / bl, i};
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      if (!mp[y]) mp[y] = ++now;
      y = mp[y], c[i] = {s[x], y}, s[x] = y, pos[i] = x;
    }
  }
  sort(b + 1, b + Q + 1, cmp1);
  vis[0] = 1e9;
  b[0] = {1, 1, 0, 0, 0, 0};
  for (int i = 1; i <= Q; i++) {
    while (b[i - 1].l > b[i].l) add(a[--b[i - 1].l]);
    while (b[i - 1].r < b[i].r) add(a[++b[i - 1].r]);
    while (b[i - 1].l < b[i].l) del(a[b[i - 1].l++]);
    while (b[i - 1].r > b[i].r) del(a[b[i - 1].r--]);
    while (b[i - 1].t < b[i].t) change(++b[i - 1].t, b[i - 1].l, b[i - 1].r);
    while (b[i - 1].t > b[i].t) undo(b[i - 1].t--, b[i - 1].l, b[i - 1].r);
    for (int j = 1; j <= n; j++)
      if (!vis[j]) {
        b[i].ans = j;
        break;
      }
  }
  sort(b + 1, b + Q + 1, cmp2);
  for (int i = 1; i <= Q; i++) printf("%d\n", b[i].ans);
  return 0;
}