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 int maxn = 2e5 + 10;
int n, m, tim, L, R, tot, sz, qsz;
int sum[maxn], cnt[maxn], ans[maxn], now[maxn];
int a[maxn], b[maxn];
struct node {
  int l, r, id, t;
  node() {}
  node(int l, int r, int id, int t) : l(l), r(r), id(id), t(t) {}
} qu[maxn];
struct node1 {
  int pos, x, y;
  node1() {}
  node1(int pos, int x, int y) : pos(pos), x(x), y(y) {}
} c[maxn];
int cmp(node a, node b) {
  if (a.l / sz == b.l / sz) {
    if (a.r / sz == b.r / sz) return a.t < b.t;
    return a.r < b.r;
  }
  return a.l < b.l;
}
void add(int val) {
  cnt[sum[val]]--;
  sum[val]++;
  cnt[sum[val]]++;
}
void del(int val) {
  cnt[sum[val]]--;
  sum[val]--;
  cnt[sum[val]]++;
}
void change(int pos, int x) {
  if (L <= pos && pos <= R) {
    del(now[pos]);
    add(x);
  }
  now[pos] = x;
}
int main() {
  scanf("%d%d", &n, &m);
  sz = (int)pow(n, 0.66666667);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[++tot] = a[i];
    now[i] = a[i];
  }
  qsz = tim = 0;
  for (int i = 1; i <= m; i++) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      qu[++qsz] = node(l, r, qsz, tim);
    } else {
      int pos, x;
      scanf("%d%d", &pos, &x);
      b[++tot] = x;
      c[++tim] = node1(pos, x, now[pos]);
      now[pos] = x;
    }
  }
  sort(qu + 1, qu + qsz + 1, cmp);
  sort(b + 1, b + tot + 1);
  tot = unique(b + 1, b + tot + 1) - b;
  for (int i = 1; i <= n; i++)
    now[i] = lower_bound(b + 1, b + tot + 1, a[i]) - b;
  for (int i = 1; i <= tim; i++) {
    c[i].x = lower_bound(b + 1, b + tot + 1, c[i].x) - b;
    c[i].y = lower_bound(b + 1, b + tot + 1, c[i].y) - b;
  }
  tim = 0;
  for (int i = 1; i <= qsz; i++) {
    while (L > qu[i].l) add(now[--L]);
    while (R < qu[i].r) add(now[++R]);
    while (L < qu[i].l) del(now[L++]);
    while (R > qu[i].r) del(now[R--]);
    while (tim < qu[i].t) tim++, change(c[tim].pos, c[tim].x);
    while (tim > qu[i].t) change(c[tim].pos, c[tim].y), tim--;
    int mex = 1;
    while (cnt[mex] > 0) mex++;
    ans[qu[i].id] = mex;
  }
  for (int i = 1; i <= qsz; i++) printf("%d\n", ans[i]);
  return 0;
}