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>
#pragma GCC optimize(3)
char c;
long long sum;
int n, m, id, l = 1, r, nq, nr, nw, block;
long long bj[500005], ans[500005];
long long tot[500005], qe[500005], cnt;
struct xin {
  int color;
} a[500005];
struct quest {
  int l, r, id, fixn;
} q[500005];
struct fix {
  int number, color;
} f[500005];
inline int in() {
  int sum = 0;
  int sign = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + ch - '0';
    ch = getchar();
  }
  return sum * sign;
}
inline void add(int x) {
  tot[bj[x]]--;
  tot[++bj[x]]++;
}
inline void del(int x) {
  tot[bj[x]]--;
  tot[--bj[x]]++;
}
inline bool cmt(quest x, quest y) {
  if (x.l / block == y.l / block) {
    if (x.r / block == y.r / block) {
      if (x.r / block & 1) return x.fixn > y.fixn;
      return x.fixn < y.fixn;
    }
    return x.r < y.r;
  }
  return x.l / block < y.l / block;
}
inline void work(int v, int i) {
  if (f[v].number >= q[i].l && f[v].number <= q[i].r) {
    del(a[f[v].number].color);
    add(f[v].color);
  }
  std::swap(a[f[v].number].color, f[v].color);
}
inline void write(int x) {
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
int main() {
  n = in();
  m = in();
  block = std::pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) {
    a[i].color = in();
    qe[++cnt] = a[i].color;
  }
  for (int i = 1, x, y; i <= m; i++) {
    c = in();
    x = in();
    y = in();
    if (c == 1) {
      q[++nq].l = x;
      q[nq].r = y;
      q[nq].id = nq;
      q[nq].fixn = nr;
    } else {
      f[++nr].number = x;
      f[nr].color = y;
      qe[++cnt] = f[nr].color;
    }
  }
  std::sort(qe + 1, qe + cnt + 1);
  cnt = std::unique(qe + 1, qe + cnt + 1) - (qe + 1);
  for (int i = 1; i <= n; i++) {
    a[i].color = std::lower_bound(qe + 1, qe + cnt + 1, a[i].color) - qe;
  }
  for (int i = 1; i <= nr; i++) {
    f[i].color = std::lower_bound(qe + 1, qe + cnt + 1, f[i].color) - qe;
  }
  std::sort(q + 1, q + nq + 1, cmt);
  for (int i = 1; i <= nq; i++) {
    while (r < q[i].r) r++, add(a[r].color);
    while (l > q[i].l) l--, add(a[l].color);
    while (r > q[i].r) del(a[r].color), r--;
    while (l < q[i].l) del(a[l].color), l++;
    while (nw < q[i].fixn) work(++nw, i);
    while (nw > q[i].fixn) work(nw--, i);
    for (ans[q[i].id] = 1; tot[ans[q[i].id]]; ans[q[i].id]++)
      ;
  }
  for (int i = 1; i <= nq; i++) write(ans[i]), putchar('\n');
}