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 maxn = 1e6 + 7;
int b[2 * maxn], a[maxn], pos[maxn], m = 0, cnt2[2 * maxn], cnt1[maxn],
                                     ans[maxn];
int l = 1, r = 0, now = 0;
struct Query {
  int l, r, pre, id;
} q[maxn];
struct Modify {
  int p, x;
} c[maxn];
bool cmp(Query a, Query b) {
  if (pos[a.l] != pos[b.l]) return a.l < b.l;
  if (pos[a.r] != pos[b.r]) return a.r < b.r;
  return a.id < b.id;
}
void add(int i) {
  cnt1[cnt2[a[i]]++]--;
  cnt1[cnt2[a[i]]]++;
}
void del(int i) {
  cnt1[cnt2[a[i]]--]--;
  cnt1[cnt2[a[i]]]++;
}
void modify(int i) {
  if (c[i].p < l || c[i].p > r) {
    swap(a[c[i].p], c[i].x);
    return;
  }
  int p1 = a[c[i].p], p2 = c[i].x;
  cnt1[cnt2[p1]--]--;
  cnt1[cnt2[p1]]++;
  cnt1[cnt2[p2]++]--;
  cnt1[cnt2[p2]]++;
  swap(a[c[i].p], c[i].x);
}
int main() {
  int n, Q, cnt = 0, op, pre = 0, tol = 0;
  scanf("%d%d", &n, &Q);
  int sz = (int)pow(n, 0.667);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[i] = (i - 1) / sz + 1;
    b[tol++] = a[i];
  }
  for (int i = 0; i < Q; i++) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &q[cnt].l, &q[cnt].r);
      q[cnt].id = cnt;
      q[cnt++].pre = pre;
    } else {
      pre++;
      scanf("%d%d", &c[pre].p, &c[pre].x);
      b[tol++] = c[pre].x;
    }
  }
  sort(b, b + tol);
  m = 1;
  for (int i = 1; i < tol; i++) {
    if (b[i] != b[i - 1]) b[m++] = b[i];
  }
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(b, b + m, a[i]) - b;
  }
  for (int i = 1; i <= pre; i++) {
    c[i].x = lower_bound(b, b + m, c[i].x) - b;
  }
  sort(q, q + cnt, cmp);
  for (int i = 0; i < cnt; i++) {
    while (r < q[i].r) {
      r++;
      add(r);
    }
    while (l > q[i].l) {
      l--;
      add(l);
    }
    while (l < q[i].l) {
      del(l);
      l++;
    }
    while (r > q[i].r) {
      del(r);
      r--;
    }
    while (now < q[i].pre) {
      now++;
      modify(now);
    }
    while (now > q[i].pre) {
      modify(now);
      now--;
    }
    for (int j = 1;; j++) {
      if (cnt1[j] == 0) {
        ans[q[i].id] = j;
        break;
      }
    }
  }
  for (int i = 0; i < cnt; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}