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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
struct Query {
  int x, y, z, idx;
};
const int N = 1e5 + 10;
const int M = 500;
int v[N], w[N], cnt[N << 1];
int mex[N];
inline void update(int x, int delta) {
  if (cnt[x] > 0 && cnt[x] < M) {
    --mex[cnt[x]];
  }
  cnt[x] += delta;
  if (cnt[x] > 0 && cnt[x] < M) {
    ++mex[cnt[x]];
  }
}
void advance(const Query& t, int sign, pair<int, int> range) {
  int x = t.x, y = t.y, z = t.z;
  if (sign > 0) {
    if (x >= range.first && x < range.second) {
      update(v[x], -1);
      update(z, 1);
    }
    v[x] = z;
  } else {
    if (x >= range.first && x < range.second) {
      update(v[x], -1);
      update(y, 1);
    }
    v[x] = y;
  }
}
void move(int t, int sign) {
  if (sign > 0) {
    update(v[t], 1);
  } else {
    update(v[t], -1);
  }
}
map<int, int> V;
int index(int x) {
  if (V.count(x)) return V[x];
  return V[x] = V.size();
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &v[i]);
    w[i] = v[i] = index(v[i]);
  }
  vector<Query> b, c;
  for (int i = 0; i < m; ++i) {
    int t, x, y;
    scanf("%d%d%d", &t, &x, &y);
    if (t == 1) {
      --x;
      b.push_back({x, y, (int)c.size(), (int)b.size()});
    } else {
      --x;
      y = index(y);
      c.push_back({x, w[x], y, -1});
      w[x] = y;
    }
  }
  int bulk = (int)(pow((double)n, 2.0 / 3) + 1);
  sort(b.begin(), b.end(), [bulk](const Query& a, const Query& b) {
    int ax = a.x / bulk, bx = b.x / bulk;
    int ay = a.y / bulk, by = b.y / bulk;
    return ax < bx || (ax == bx && ay < by) ||
           (ax == bx && ay == by && a.z < b.z);
  });
  int x = 0, y = 0, z = 0;
  vector<int> ret(b.size());
  for (auto& it : b) {
    while (z > it.z) advance(c[--z], -1, {x, y});
    while (z < it.z) advance(c[z++], 1, {x, y});
    while (x > it.x) move(--x, 1);
    while (x < it.x) move(x++, -1);
    while (y > it.y) move(--y, -1);
    while (y < it.y) move(y++, 1);
    for (int i = 1; i < M; ++i) {
      if (!mex[i]) {
        ret[it.idx] = i;
        break;
      }
    }
  }
  for (int i = 0; i < ret.size(); ++i) {
    printf("%d\n", ret[i]);
  }
  return 0;
}