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;
map<int, int> ma;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
int dd;
struct Q {
  int l, r, t, id;
} q[400010];
struct C {
  int pos, x, o;
} c[400010];
int block;
bool comp(Q u, Q v) {
  return u.l / block < v.l / block ||
         u.l / block == v.l / block && u.r / block < v.r / block ||
         u.l / block == v.l / block && u.r / block == v.r / block && u.t < v.t;
}
int a[400010];
int b[400010];
int n, m;
int d[400010];
int e[400010];
void add(int x, int y) {
  b[e[x]]--;
  e[x] += y;
  b[e[x]]++;
}
int ans[400010];
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (ma.count(a[i]))
      a[i] = ma[a[i]];
    else
      ma[a[i]] = ++dd, a[i] = dd;
    d[i] = a[i];
  }
  block = ceil(pow(n, 2. / 3));
  int cnt = 0;
  int cnt1 = 0;
  int x, y, z;
  for (int i = 1; i <= m; i++) {
    x = read(), y = read(), z = read();
    if (x == 1)
      q[++cnt].l = y, q[cnt].r = z, q[cnt].t = cnt1, q[cnt].id = cnt;
    else {
      if (ma.count(z))
        z = ma[z];
      else
        ma[z] = ++dd, z = dd;
      c[++cnt1].pos = y, c[cnt1].x = z, c[cnt1].o = d[y];
      d[y] = z;
    }
  }
  int l, r, t;
  l = r = t = 0;
  sort(q + 1, q + cnt + 1, comp);
  for (int i = 1; i <= cnt; i++) {
    while (r < q[i].r) {
      r++;
      add(a[r], 1);
    }
    while (l > q[i].l) {
      l--;
      add(a[l], 1);
    }
    while (r > q[i].r) {
      if (r <= n) add(a[r], -1);
      r--;
    }
    while (l < q[i].l) {
      if (l) add(a[l], -1);
      l++;
    }
    while (t < q[i].t) {
      ++t;
      int pos = c[t].pos;
      a[pos] = c[t].x;
      if (pos >= l && pos <= r) add(c[t].o, -1), add(c[t].x, 1);
    }
    while (t > q[i].t) {
      int pos = c[t].pos;
      a[pos] = c[t].o;
      if (pos >= l && pos <= r) add(c[t].o, 1), add(c[t].x, -1);
      t--;
    }
    for (int j = 1; j <= 1000; j++) {
      if (!b[j]) {
        ans[q[i].id] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= cnt; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}