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 N = 1e5 + 5, M = 1 << 9;
int A[N], AA[N], TP[N], L[N], R[N], AH[N], T[N << 1], TT[N << 1], K[N], RT[N],
    blk, n, q, nn, ll = 1, rr = 1, tt;
vector<int> v;
struct node {
  int l, r, t;
} Q[N];
bool cmp(const node& a, const node& b) {
  if (AH[a.l] ^ AH[b.l]) return (a.l < b.l);
  if (AH[a.r] ^ AH[b.r]) return (a.r < b.r);
  return (a.t < b.t);
}
void addmo(int ind, int d) {
  TT[T[A[ind]]]--;
  T[A[ind]] += d;
  TT[T[A[ind]]]++;
}
int main() {
  scanf("%d %d", &n, &q);
  blk = pow(n, 2.1 / 3.0);
  for (int i = 0; i < N; i++) AH[i] = i / blk;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
    v.push_back(A[i]);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d %d %d", &TP[i], &L[i], &R[i]);
    if (TP[i] == 2) v.push_back(R[i]);
  }
  set<int> st(v.begin(), v.end());
  v.assign(st.begin(), st.end());
  for (int i = 1; i <= n; i++)
    A[i] = lower_bound(v.begin(), v.end(), A[i]) - v.begin();
  for (int i = 1; i <= q; i++)
    if (TP[i] == 2) R[i] = lower_bound(v.begin(), v.end(), R[i]) - v.begin();
  memcpy(AA, A, sizeof(AA));
  for (int i = 1; i <= q; i++)
    if (TP[i] == 2) {
      K[i] = AA[L[i]];
      AA[L[i]] = R[i];
    }
  for (int i = 1; i <= q; i++)
    if (TP[i] == 1) Q[++nn] = {L[i], R[i], i};
  sort(Q + 1, Q + nn + 1, cmp);
  T[A[1]]++;
  TT[1]++;
  for (int i = 1; i <= nn; i++) {
    while (ll > Q[i].l) addmo(--ll, 1);
    while (rr < Q[i].r) addmo(++rr, 1);
    while (ll < Q[i].l) addmo(ll++, -1);
    while (rr > Q[i].r) addmo(rr--, -1);
    while (tt < Q[i].t) {
      tt++;
      if (TP[tt] == 1) continue;
      if (ll <= L[tt] && L[tt] <= rr) {
        addmo(L[tt], -1);
        A[L[tt]] = R[tt];
        addmo(L[tt], 1);
        continue;
      }
      A[L[tt]] = R[tt];
    }
    while (tt > Q[i].t) {
      if (TP[tt] == 1) {
        tt--;
        continue;
      }
      if (ll <= L[tt] && L[tt] <= rr) {
        addmo(L[tt], -1);
        A[L[tt]] = K[tt];
        addmo(L[tt], 1);
        tt--;
        continue;
      }
      A[L[tt]] = K[tt];
      tt--;
    }
    for (int j = 1;; j++)
      if (!TT[j]) {
        RT[Q[i].t] = j;
        break;
      }
  }
  for (int i = 1; i <= q; i++)
    if (TP[i] == 1) printf("%d\n", RT[i]);
}