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 = 300005;
int a[N], e[N], f[N];
int k, c[N], d[N], b[N];
map<int, int> ma;
struct que {
  int l, r, posl, posr, ts, id;
} q[N];
struct modi {
  int pos, x, pre;
} mo[N];
bool cmp(que a, que b) {
  if (a.posl == b.posl) {
    if (a.posr == b.posr) return a.ts < b.ts;
    return a.posr < b.posr;
  }
  return a.posl < b.posl;
}
void update(int x, int p) {
  if (x == 0) return;
  if (c[x] > 0) d[c[x]]--;
  c[x] += p;
  if (c[x] > 0) d[c[x]]++;
}
void modify(int x, int p, int l, int r) {
  if (x >= l && x <= r) {
    update(a[x], -1);
    update(p, 1);
  }
  a[x] = p;
}
int fun() {
  for (int i = 1;; i++)
    if (d[i] < 1) return i;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int p = pow(n, 0.666666);
  if (p == 0) p = 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), e[i] = f[i] = a[i];
  sort(f + 1, f + 1 + n);
  k = unique(f + 1, f + 1 + n) - (f + 1);
  for (int i = 1; i <= k; i++) ma[f[i]] = i;
  for (int i = 1; i <= n; i++) e[i] = a[i] = ma[a[i]];
  int tot = 0, cnt = 0, len = k;
  int op, x, y;
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      q[tot].l = x, q[tot].posl = x / p;
      q[tot].r = y, q[tot].posr = y / p;
      q[tot].id = tot, q[tot++].ts = cnt;
    } else {
      if (!ma[y]) ma[y] = ++len;
      y = ma[y];
      mo[++cnt].pos = x, mo[cnt].pre = e[x];
      mo[cnt].x = y, e[x] = y;
    }
  }
  sort(q, q + tot, cmp);
  int l = 0, r = 0, cn = 0;
  for (int i = 0; i < tot; i++) {
    while (cn < q[i].ts) modify(mo[cn + 1].pos, mo[cn + 1].x, l, r), cn++;
    while (cn > q[i].ts) modify(mo[cn].pos, mo[cn].pre, l, r), cn--;
    while (l < q[i].l) update(a[l++], -1);
    while (l > q[i].l) update(a[--l], 1);
    while (r > q[i].r) update(a[r--], -1);
    while (r < q[i].r) update(a[++r], 1);
    b[q[i].id] = fun();
  }
  for (int i = 0; i < tot; i++) printf("%d\n", b[i]);
  return 0;
}