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;
struct node {
  int l, r, last, id;
} q[1001000];
struct node1 {
  int x, old, now;
} modi[1001000];
int pos[1001000], vis[1001000], minn[1001000], num[1001000], cnt, cnt1, n, l, r,
    o, p, t, a[1001000];
int size, ex[1001000], excnt, i, ans[1001000], op, bl, m, b[1001000];
inline int read() {
  int x = 0, y = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') y = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * y;
}
inline bool cmp(node x, node y) {
  if (pos[x.l] == pos[y.l])
    if (pos[x.r] == pos[y.r])
      return x.last < y.last;
    else
      return x.r < y.r;
  else
    return x.l < y.l;
}
inline void add(int x, int f) {
  vis[num[x]]--;
  num[x] += f;
  vis[num[x]]++;
}
inline void init() {
  sort(ex + 1, ex + 1 + excnt);
  size = unique(ex + 1, ex + 1 + excnt) - ex - 1;
  for (i = 1; i <= n; i++) a[i] = lower_bound(ex + 1, ex + 1 + size, a[i]) - ex;
  for (i = 1; i <= cnt1; i++) {
    modi[i].old = lower_bound(ex + 1, ex + 1 + size, modi[i].old) - ex;
    modi[i].now = lower_bound(ex + 1, ex + 1 + size, modi[i].now) - ex;
  }
}
void solve() {
  vis[0] = 1000000000;
  l = 1;
  for (i = 1; i <= cnt; i++) {
    while (l > q[i].l) l--, add(a[l], 1);
    while (r < q[i].r) r++, add(a[r], 1);
    while (l < q[i].l) add(a[l], -1), l++;
    while (r > q[i].r) add(a[r], -1), r--;
    while (t < q[i].last) {
      t++;
      if (modi[t].x >= l && modi[t].x <= r) add(a[modi[t].x], -1);
      a[modi[t].x] = modi[t].now;
      if (modi[t].x >= l && modi[t].x <= r) add(a[modi[t].x], 1);
    }
    while (t > q[i].last) {
      if (modi[t].x >= l && modi[t].x <= r) add(a[modi[t].x], -1);
      a[modi[t].x] = modi[t].old;
      if (modi[t].x >= l && modi[t].x <= r) add(a[modi[t].x], 1);
      t--;
    }
    while (vis[ans[q[i].id]]) ans[q[i].id]++;
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; i++) {
    a[i] = read();
    ex[++excnt] = a[i];
    b[i] = a[i];
  }
  while (m--) {
    op = read(), o = read(), p = read();
    if (op == 1) {
      q[++cnt].l = o, q[cnt].r = p, q[cnt].id = cnt, q[cnt].last = cnt1;
    } else {
      modi[++cnt1].x = o, modi[cnt1].old = b[o];
      modi[cnt1].now = p;
      b[o] = p;
      ex[++excnt] = p;
    }
  }
  init();
  bl = 2000;
  for (i = 1; i <= n; i++) pos[i] = (i - 1) / bl + 1;
  sort(q + 1, q + 1 + cnt, cmp);
  solve();
  for (i = 1; i <= cnt; i++) printf("%d\n", ans[i]);
}