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 = 500001;
int c[maxn], unit[maxn], n, m;
int ans[maxn], vis[maxn];
long long A[maxn], sum;
int block, cntq, cntm, tot;
std::map<int, int> mp;
struct Q {
  int l, r, id, t;
} query[maxn];
struct M {
  int id, x, y;
} modify[maxn];
bool cmp(Q a, Q b) {
  if (unit[a.l] == unit[b.l]) {
    if (unit[a.r] == unit[b.r]) return a.t < b.t;
    return a.r < b.r;
  } else
    return a.l < b.l;
}
void update(int p, int val) {
  vis[A[c[p]]]--;
  A[c[p]] += val;
  vis[A[c[p]]]++;
}
void work() {
  int left = 1, right = 0, cntmdy = 0;
  for (int i = 1; i <= cntq; i++) {
    while (right < query[i].r) {
      update(right + 1, 1);
      right++;
    }
    while (right > query[i].r) {
      update(right, -1);
      right--;
    }
    while (left < query[i].l) {
      update(left, -1);
      left++;
    }
    while (left > query[i].l) {
      update(left - 1, 1);
      left--;
    }
    while (cntmdy < query[i].t) {
      cntmdy++;
      if (left <= modify[cntmdy].x && modify[cntmdy].x <= right) {
        update(modify[cntmdy].x, -1);
      }
      swap(c[modify[cntmdy].x], modify[cntmdy].y);
      if (left <= modify[cntmdy].x && modify[cntmdy].x <= right) {
        update(modify[cntmdy].x, 1);
      }
    }
    while (cntmdy > query[i].t) {
      if (left <= modify[cntmdy].x && modify[cntmdy].x <= right) {
        update(modify[cntmdy].x, -1);
      }
      swap(c[modify[cntmdy].x], modify[cntmdy].y);
      if (left <= modify[cntmdy].x && modify[cntmdy].x <= right) {
        update(modify[cntmdy].x, 1);
      }
      cntmdy--;
    }
    int ret = 1;
    while (vis[ret]) ret++;
    ans[query[i].id] = ret;
  }
}
int main(int argc, char const *argv[]) {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    if (mp[tmp])
      c[i] = mp[tmp];
    else
      c[i] = mp[tmp] = ++tot;
  }
  int block = (int)pow(n, 2.0 / 3.0);
  for (int i = 1; i <= n; i++) {
    unit[i] = (i - 1) / block + 1;
  }
  for (int i = 1; i <= m; i++) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      cntq++;
      query[cntq].l = x;
      query[cntq].r = y;
      query[cntq].id = cntq;
      query[cntq].t = cntm;
    } else {
      cntm++;
      if (mp[y])
        y = mp[y];
      else
        y = mp[y] = ++tot;
      modify[cntm].x = x;
      modify[cntm].y = y;
    }
  }
  sort(query + 1, query + cntq + 1, cmp);
  work();
  for (int i = 1; i <= cntq; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}