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 MAX_N = 300024;
int pos[MAX_N], col[MAX_N], b[MAX_N], ans[MAX_N], Num[MAX_N], flag[MAX_N],
    lst[MAX_N];
struct node {
  int l, r, pos, k;
} q[MAX_N];
struct opti {
  int x, y, z;
} o[MAX_N];
int L = 1, R = 0, Ans = 0, K = 0, q_num = 0, o_num;
bool cmp(const node &a, const node &b) {
  if (pos[a.l] ^ pos[b.l]) return pos[a.l] < pos[b.l];
  if (pos[a.r] ^ pos[b.r]) return pos[a.r] < pos[b.r];
  return a.k < b.k;
}
void add(int x) {
  Num[flag[col[x]]]--;
  Num[++flag[col[x]]]++;
}
void del(int x) {
  Num[flag[col[x]]]--;
  Num[--flag[col[x]]]++;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int blo = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &col[i]);
    pos[i] = i / blo;
    b[i] = col[i];
    lst[i] = col[i];
  }
  for (int i = 1; i <= m; ++i) {
    int opt, x, y;
    scanf("%d%d%d", &opt, &x, &y);
    if (opt == 1) {
      q[++q_num].l = x, q[q_num].r = y, q[q_num].pos = q_num,
      q[q_num].k = o_num;
    } else {
      o[++o_num].x = x, o[o_num].y = y, o[o_num].z = lst[x], lst[x] = y,
      b[n + o_num] = y;
    }
  }
  sort(b + 1, b + 1 + n + o_num);
  int b_num = unique(b + 1, b + 1 + n + o_num) - b - 1;
  sort(q + 1, q + 1 + q_num, cmp);
  for (int i = 1; i <= n; i++) {
    col[i] = lower_bound(b + 1, b + 1 + b_num, col[i]) - b;
  }
  for (int i = 1; i <= o_num; i++) {
    o[i].z = lower_bound(b + 1, b + 1 + b_num, o[i].z) - b;
    o[i].y = lower_bound(b + 1, b + 1 + b_num, o[i].y) - b;
  }
  for (int i = 1; i <= q_num; i++) {
    while (K < q[i].k) {
      ++K;
      if (o[K].x <= R && o[K].x >= L) {
        Num[flag[col[o[K].x]]]--;
        Num[--flag[col[o[K].x]]]++;
        Num[flag[o[K].y]]--;
        Num[++flag[o[K].y]]++;
      }
      col[o[K].x] = o[K].y;
    }
    while (K > q[i].k) {
      if (o[K].x <= R && o[K].x >= L) {
        Num[flag[col[o[K].x]]]--;
        Num[--flag[col[o[K].x]]]++;
        Num[flag[o[K].z]]--;
        Num[++flag[o[K].z]]++;
      }
      col[o[K].x] = o[K].z;
      --K;
    }
    while (R < q[i].r) {
      R++;
      add(R);
    }
    while (L > q[i].l) {
      L--;
      add(L);
    }
    while (L < q[i].l) {
      del(L);
      L++;
    }
    while (R > q[i].r) {
      del(R);
      R--;
    }
    for (int i = 1; i <= 2000000; i++)
      if (!Num[i]) {
        Ans = i;
        break;
      }
    ans[q[i].pos] = Ans;
  }
  for (int i = 1; i <= q_num; ++i) printf("%d\n", ans[i]);
  return 0;
}