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;
inline int read() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(register int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 100005;
const int M = 100005;
int n, m, nown, cnt_q, cnt_r, T;
int pos[M], col[M], ans[M];
int cnt[2 * N], sum[2 * N], a[N], lsh[N * 2];
map<int, int> mp;
struct node {
  int tim, pos, l, r;
} q[M];
int cmp(node a, node b) {
  if (a.l / T != b.l / T) {
    return a.l < b.l;
  } else if (a.r / T != b.r / T) {
    return a.r < b.r;
  } else
    return a.tim < b.tim;
}
void Add(int x) {
  sum[cnt[a[x]]]--;
  cnt[a[x]]++;
  sum[cnt[a[x]]]++;
}
void Delete(int x) {
  sum[cnt[a[x]]]--;
  cnt[a[x]]--;
  sum[cnt[a[x]]]++;
}
void modify(int x, int t) {
  if (pos[t] >= q[x].l && pos[t] <= q[x].r) {
    Delete(pos[t]);
    int tmp = a[pos[t]];
    a[pos[t]] = col[t];
    Add(pos[t]);
    a[pos[t]] = tmp;
  }
  swap(a[pos[t]], col[t]);
}
int find() {
  for (int i = 1; i <= n; i++) {
    if (sum[i] == 0) return i;
  }
}
void work() {
  sort(lsh, lsh + nown);
  int size = unique(lsh, lsh + nown) - lsh;
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(lsh, lsh + size, a[i]) - lsh;
  }
  for (int i = 1; i <= cnt_r; i++) {
    col[i] = lower_bound(lsh, lsh + size, col[i]) - lsh;
  }
  return;
}
int main() {
  n = read();
  m = read();
  T = pow(n, 0.6666);
  for (int i = 1; i <= n; i++) a[i] = lsh[nown++] = read();
  for (int i = 1; i <= m; i++) {
    int t = read();
    if (t == 1) {
      cnt_q++;
      q[cnt_q].tim = cnt_r;
      q[cnt_q].pos = cnt_q;
      cin >> q[cnt_q].l >> q[cnt_q].r;
    } else {
      cnt_r++;
      cin >> pos[cnt_r] >> col[cnt_r];
      lsh[nown++] = col[cnt_r];
    }
  }
  work();
  sort(q + 1, q + cnt_q + 1, cmp);
  int cur = 0, x = 1, y = 0;
  for (int i = 1; i <= cnt_q; i++) {
    while (x < q[i].l) {
      Delete(x);
      x++;
    }
    while (x > q[i].l) {
      x--;
      Add(x);
    }
    while (y < q[i].r) {
      y++;
      Add(y);
    }
    while (y > q[i].r) {
      Delete(y);
      y--;
    }
    while (cur < q[i].tim) {
      cur++;
      modify(i, cur);
    }
    while (cur > q[i].tim) {
      modify(i, cur);
      cur--;
    }
    ans[q[i].pos] = find();
  }
  for (int i = 1; i <= cnt_q; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}