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 = 200010;
struct node {
  int l, r, id, ans, tim;
} q[N];
struct nodee {
  int x, y, z;
} b[N];
int n, m, pos[N], L[N], R[N], a[N], qn, f[N], h[N], g[N], l, r, cnt, c[N];
int read() {
  int d = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') d = d * 10 + c - 48, c = getchar();
  return d * f;
}
bool cmp1(node a, node b) {
  return pos[a.l] < pos[b.l] || pos[a.l] == pos[b.l] && pos[a.r] < pos[b.r] ||
         pos[a.l] == pos[b.l] && pos[a.r] == pos[b.r] && a.tim < b.tim;
}
bool cmp2(node a, node b) { return a.id < b.id; }
void add(int x) {
  if (--f[h[a[x]]] == 0) g[pos[h[a[x]]]]--;
  h[a[x]]++;
  if (++f[h[a[x]]] == 1) g[pos[h[a[x]]]]++;
}
void del(int x) {
  if (--f[h[a[x]]] == 0) g[pos[h[a[x]]]]--;
  h[a[x]]--;
  if (++f[h[a[x]]] == 1) g[pos[h[a[x]]]]++;
}
void add_time(int k) {
  int x = b[k].x, y = b[k].y, z = b[k].z;
  if (x >= l && x <= r) del(x);
  b[k].z = a[x];
  a[x] = y;
  if (x >= l && x <= r) add(x);
}
void del_time(int k) {
  int x = b[k].x, y = b[k].y, z = b[k].z;
  if (x >= l && x <= r) del(x);
  a[x] = z;
  if (x >= l && x <= r) add(x);
}
int query() {
  int i = 1;
  for (int i = 1; i <= pos[n]; i++)
    if (g[i] != R[i] - L[i] + 1) break;
  for (int j = L[i]; j <= R[i]; j++)
    if (!f[j]) return j;
}
int main() {
  n = read();
  m = read();
  int blo = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) {
    pos[i] = (i - 1) / blo + 1;
    if (!L[pos[i]]) L[pos[i]] = i;
    R[pos[i]] = i;
  }
  for (int i = 1; i <= n; i++) a[i] = read(), c[++cnt] = a[i];
  int now = 0;
  for (int i = 1; i <= m; i++) {
    int op = read(), l = read(), r = read();
    if (op == 1)
      q[++qn] = (node){l, r, i, 0, now};
    else
      b[++now] = (nodee){l, r, 0}, c[++cnt] = b[now].y;
  }
  sort(c + 1, c + 1 + cnt);
  cnt = unique(c + 1, c + cnt + 1) - c - 1;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(c + 1, c + 1 + cnt, a[i]) - c;
  for (int i = 1; i <= now; i++)
    b[i].y = lower_bound(c + 1, c + 1 + cnt, b[i].y) - c;
  sort(q + 1, q + 1 + qn, cmp1);
  l = 1;
  r = 0;
  now = 0;
  for (int i = 1; i <= qn; i++) {
    while (r < q[i].r) add(++r);
    while (r > q[i].r) del(r--);
    while (l > q[i].l) add(--l);
    while (l < q[i].l) del(l++);
    while (now < q[i].tim) add_time(++now);
    while (now > q[i].tim) del_time(now--);
    q[i].ans = query();
  }
  sort(q + 1, q + 1 + qn, cmp2);
  for (int i = 1; i <= qn; i++) printf("%d\n", q[i].ans);
  return 0;
}