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 MOD = 10007;
const int N = 200000 + 5;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
using namespace std;
struct Node {
  int l, r;
  int time;
  int id;
} q[N];
struct Change {
  int pos;
  int col;
} c[N];
int n, m, a[N], temp[N];
int block;
int numQ, numC;
int cnt[N], sum[N];
int res[N];
bool cmp(Node a, Node b) {
  return (a.l / block) ^ (b.l / block)
             ? a.l < b.l
             : ((a.r / block) ^ (b.r / block) ? a.r < b.r : a.time < b.time);
}
void add(int x) {
  sum[cnt[x]]--;
  cnt[x]++;
  sum[cnt[x]]++;
}
void del(int x) {
  sum[cnt[x]]--;
  cnt[x]--;
  sum[cnt[x]]++;
}
void change(int x, int ti) {
  if (c[ti].pos >= q[x].l && c[ti].pos <= q[x].r) {
    del(a[c[ti].pos]);
    add(c[ti].col);
  }
  swap(c[ti].col, a[c[ti].pos]);
}
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    numQ = 0;
    numC = 0;
    memset(sum, 0, sizeof(sum));
    memset(cnt, 0, sizeof(cnt));
    block = pow(n, 0.66666);
    int numTemp = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      temp[++numTemp] = a[i];
    }
    for (int i = 1; i <= m; i++) {
      int op;
      scanf("%d", &op);
      if (op == 1) {
        ++numQ;
        scanf("%d%d", &q[numQ].l, &q[numQ].r);
        q[numQ].id = numQ;
        q[numQ].time = numC;
      } else {
        ++numC;
        scanf("%d%d", &c[numC].pos, &c[numC].col);
        temp[++numTemp] = c[numC].col;
      }
    }
    sort(temp + 1, temp + 1 + numTemp);
    numTemp = unique(temp + 1, temp + 1 + numTemp) - temp;
    for (int i = 1; i <= n; i++)
      a[i] = lower_bound(temp + 1, temp + 1 + numTemp, a[i]) - temp;
    for (int i = 1; i <= numC; i++)
      c[i].col = lower_bound(temp + 1, temp + 1 + numTemp, c[i].col) - temp;
    sort(q + 1, q + numQ + 1, cmp);
    int l = 1, r = 0, time = 0;
    for (int i = 1; i <= numQ; i++) {
      int ql = q[i].l, qr = q[i].r;
      int qtime = q[i].time;
      while (l > ql) add(a[--l]);
      while (l < ql) del(a[l++]);
      while (r < qr) add(a[++r]);
      while (r > qr) del(a[r--]);
      while (time < qtime) change(i, ++time);
      while (time > qtime) change(i, time--);
      res[q[i].id] = 1;
      while (sum[res[q[i].id]] > 0) res[q[i].id]++;
    }
    for (int i = 1; i <= numQ; i++) printf("%d\n", res[i]);
  }
  return 0;
}