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 = 200009;
int cnt[maxn], sum[maxn], a[maxn], bel[maxn], ans[maxn], t[maxn], b[maxn];
bool vis[maxn];
struct Q {
  int l, r, idx, tim;
} q[maxn];
struct C {
  int x, y, lst;
} c[maxn];
int n, m, tot, lim, now, siz, cq, l, r;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
bool cmp(Q x, Q y) {
  if (bel[x.l] == bel[y.l]) {
    if (bel[x.r] == bel[y.r]) return x.tim < y.tim;
    return bel[x.r] < bel[y.r];
  }
  return bel[x.l] < bel[y.l];
}
void update(int x) {
  if (vis[x]) {
    sum[cnt[a[x]]]--;
    cnt[a[x]]--;
    sum[cnt[a[x]]]++;
    vis[x] = 0;
  } else {
    sum[cnt[a[x]]]--;
    cnt[a[x]]++;
    sum[cnt[a[x]]]++;
    vis[x] = 1;
  }
}
void change(int idx, int opt) {
  if (opt == 1) {
    if (vis[c[idx].x]) {
      update(c[idx].x);
      a[c[idx].x] = c[idx].y;
      update(c[idx].x);
    } else
      a[c[idx].x] = c[idx].y;
  } else {
    if (vis[c[idx].x]) {
      update(c[idx].x);
      a[c[idx].x] = c[idx].lst;
      update(c[idx].x);
    } else
      a[c[idx].x] = c[idx].lst;
  }
}
int main() {
  n = read();
  m = read();
  siz = pow(n, 2.0 / 3);
  for (int i = 1; i <= n; i++) b[i] = a[i] = t[++tot] = read();
  for (int i = 1; i <= m; i++) {
    int opt = read(), x = read(), y = read();
    if (opt == 1)
      cq++, q[cq] = (Q){x, y, cq, lim};
    else
      lim++, c[lim] = (C){x, y, b[x]}, t[++tot] = y, b[x] = y;
  }
  sort(t + 1, t + 1 + tot);
  tot = unique(t + 1, t + 1 + tot) - t - 1;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(t + 1, t + 1 + tot, a[i]) - t;
  for (int i = 1; i <= lim; i++) {
    c[i].lst = lower_bound(t + 1, t + 1 + tot, c[i].lst) - t;
    c[i].y = lower_bound(t + 1, t + 1 + tot, c[i].y) - t;
  }
  for (int i = 1; i <= n; i++) bel[i] = (i - 1) / siz + 1;
  sort(q + 1, q + 1 + cq, cmp);
  l = 1, r = 0;
  for (int i = 1; i <= cq; i++) {
    while (now > q[i].tim) change(now, -1), now--;
    while (now < q[i].tim) now++, change(now, 1);
    while (l > q[i].l) l--, update(l);
    while (r < q[i].r) r++, update(r);
    while (l < q[i].l) update(l), l++;
    while (r > q[i].r) update(r), r--;
    int res = 1;
    while (sum[res]) res++;
    ans[q[i].idx] = res;
  }
  for (int i = 1; i <= cq; i++) printf("%d\n", ans[i]);
  return 0;
}