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 bs;
struct query {
  int l, r, x, id;
  friend bool operator<(const query &u, const query &v) {
    if (u.l / bs != v.l / bs) return u.l / bs < v.l / bs;
    if (u.r / bs != v.r / bs) return u.r / bs < v.r / bs;
    return u.x < v.x;
  }
} Q[100000];
int qcnt;
struct modify {
  int pos, last, val;
} M[100000];
int mcnt;
int a[200005];
int last[200005];
int ans[200005];
int cnt[200005];
int scnt[200005];
bool exist[200005];
void update(int x) {
  if (exist[x]) {
    scnt[cnt[a[x]]]--;
    cnt[a[x]]--;
    scnt[cnt[a[x]]]++;
  } else {
    scnt[cnt[a[x]]]--;
    cnt[a[x]]++;
    scnt[cnt[a[x]]]++;
  }
  exist[x] ^= 1;
}
void change(int mi) {
  int x = M[mi].pos;
  if (exist[x]) {
    update(x);
    a[x] = M[mi].val;
    update(x);
  } else
    a[x] = M[mi].val;
}
void restore(int mi) {
  int x = M[mi].pos;
  if (exist[x]) {
    update(x);
    a[x] = M[mi].last;
    update(x);
  } else
    a[x] = M[mi].last;
}
unordered_map<int, int> ls;
int lscnt;
int getls(int x) {
  if (ls.find(x) == ls.end()) ls[x] = ++lscnt;
  return ls[x];
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  bs = pow(n, 2.0 / 3.0);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = getls(x);
    last[i] = a[i];
  }
  while (q--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      Q[qcnt] = (query){l, r, mcnt - 1, qcnt};
      qcnt++;
    } else {
      r = getls(r);
      M[mcnt++] = (modify){l, last[l], r};
      last[l] = r;
    }
  }
  sort(Q, Q + qcnt);
  int l = 1, r = 0;
  int mi = -1;
  for (int i = 0; i < qcnt; i++) {
    while (mi < Q[i].x) change(++mi);
    while (mi > Q[i].x) restore(mi--);
    while (r < Q[i].r) update(++r);
    while (r > Q[i].r) update(r--);
    while (l < Q[i].l) update(l++);
    while (l > Q[i].l) update(--l);
    int cur = 1;
    while (scnt[cur] > 0) cur++;
    ans[Q[i].id] = cur;
  }
  for (int i = 0; i < qcnt; i++) printf("%d\n", ans[i]);
  return 0;
}