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 = 2e5 + 5;
int n, m, t;
int a[maxn];
struct ex {
  int l, r, t, n;
};
struct opt {
  int p, x, y;
};
vector<ex> que;
vector<opt> op;
int ans[maxn];
int size;
map<int, int> ss;
int scnt;
int s[2 * maxn];
int cnt[maxn];
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
bool cmp(ex x, ex y) {
  if (x.l / size != y.l / size) {
    return x.l < y.l;
  } else {
    if (x.r / size != y.r / size) {
      return x.r < y.r;
    } else {
      return x.t < y.t;
    }
  }
}
int main() {
  int i, j;
  int o;
  int x, y;
  n = read();
  m = read();
  for (i = 1; i <= n; i++) {
    a[i] = read();
    if (!ss[a[i]]) {
      ss[a[i]] = ++scnt;
    }
    a[i] = ss[a[i]];
  }
  for (j = 1; j <= m; j++) {
    o = read();
    x = read();
    y = read();
    if (o == 1) {
      que.push_back((ex){x, y, (int)op.size(), (int)(que.size() + 1)});
    } else {
      if (!ss[y]) {
        ss[y] = ++scnt;
      }
      y = ss[y];
      op.push_back((opt){x, y});
    }
  }
  for (i = 0; i < op.size(); i++) {
    op[i].y = a[op[i].p];
    a[op[i].p] = op[i].x;
  }
  for (i = op.size() - 1; i >= 0; i--) {
    a[op[i].p] = op[i].y;
  }
  size = pow(n, 2.0 / 3.0);
  sort(que.begin(), que.end(), cmp);
  int l = 1, r = 1, t = 0;
  s[a[1]] = 1;
  cnt[1] = 1;
  for (ex tmp : que) {
    while (l > tmp.l) {
      l--;
      cnt[s[a[l]]]--;
      s[a[l]]++;
      cnt[s[a[l]]]++;
    }
    while (r < tmp.r) {
      r++;
      cnt[s[a[r]]]--;
      s[a[r]]++;
      cnt[s[a[r]]]++;
    }
    while (l < tmp.l) {
      cnt[s[a[l]]]--;
      s[a[l]]--;
      cnt[s[a[l]]]++;
      l++;
    }
    while (r > tmp.r) {
      cnt[s[a[r]]]--;
      s[a[r]]--;
      cnt[s[a[r]]]++;
      r--;
    }
    while (t < tmp.t) {
      if (l <= op[t].p and op[t].p <= r) {
        cnt[s[a[op[t].p]]]--;
        s[a[op[t].p]]--;
        cnt[s[a[op[t].p]]]++;
      }
      a[op[t].p] = op[t].x;
      if (l <= op[t].p and op[t].p <= r) {
        cnt[s[a[op[t].p]]]--;
        s[a[op[t].p]]++;
        cnt[s[a[op[t].p]]]++;
      }
      t++;
    }
    while (t > tmp.t) {
      t--;
      if (l <= op[t].p and op[t].p <= r) {
        cnt[s[a[op[t].p]]]--;
        s[a[op[t].p]]--;
        cnt[s[a[op[t].p]]]++;
      }
      a[op[t].p] = op[t].y;
      if (l <= op[t].p and op[t].p <= r) {
        cnt[s[a[op[t].p]]]--;
        s[a[op[t].p]]++;
        cnt[s[a[op[t].p]]]++;
      }
    }
    for (i = 1; i <= 500; i++) {
      if (!cnt[i]) {
        ans[tmp.n] = i;
        break;
      }
    }
  }
  for (i = 1; i <= que.size(); i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}