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 maxx = 2e5 + 10;
struct qnode {
  int l, r, t, id;
} e[maxx];
struct cnode {
  int p, pre, suc;
} c[maxx];
int a[maxx], b[maxx], ans[maxx];
int mx[maxx], sum[maxx];
int tot, siz;
int l, r, t;
map<int, int> ma;
bool cmp(qnode a, qnode b) {
  if (a.l / siz != b.l / siz) return a.l < b.l;
  if (a.r / siz != b.r / siz) return a.r < b.r;
  return a.t < b.t;
}
void add(int x, int c) {
  mx[sum[x]]--;
  sum[x] += c;
  mx[sum[x]]++;
}
void change(int x, int y) {
  if (l <= x && x <= r) {
    add(y, 1);
    add(a[x], -1);
  }
  a[x] = y;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  siz = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!ma[a[i]]) ma[a[i]] = ++tot;
    a[i] = b[i] = ma[a[i]];
  }
  int t1 = 0, t2 = 0;
  int op, x, y;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      t1++;
      e[t1].l = x, e[t1].r = y;
      e[t1].id = t1, e[t1].t = t2;
    } else {
      t2++;
      if (!ma[y]) ma[y] = ++tot;
      c[t2].p = x;
      c[t2].pre = b[x];
      c[t2].suc = b[x] = ma[y];
    }
  }
  sort(e + 1, e + 1 + t1, cmp);
  l = 1, r = 0, t = 0;
  for (int i = 1; i <= t1; i++) {
    while (l < e[i].l) add(a[l++], -1);
    while (l > e[i].l) add(a[--l], 1);
    while (r < e[i].r) add(a[++r], 1);
    while (r > e[i].r) add(a[r--], -1);
    while (t < e[i].t) t++, change(c[t].p, c[t].suc);
    while (t > e[i].t) change(c[t].p, c[t].pre), t--;
    int num = 1;
    while (mx[num]) num++;
    ans[e[i].id] = num;
  }
  for (int i = 1; i <= t1; i++) printf("%d\n", ans[i]);
  return 0;
}