Machine Learning CodeForces - 940F (带修改的莫队)

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 c_i be the number of occurrences of i ina_l: r, where a_l: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c₀, c₁, ..., c_10⁹}
2. You are given two integers p to x. Change a_p 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, ..., a_n (1 ≤ a_i ≤ 10⁹).

Each of the next q lines describes a single query.

The first type of query is described by three integers t_i = 1, l_i, r_i, where 1 ≤ l_i ≤ r_i ≤ n — the bounds of the subarray.

The second type of query is described by three integers t_i = 2, p_i, x_i, where 1 ≤ p_i ≤ n is the index of the element, which must be changed and 1 ≤ x_i ≤ 10⁹ is the new value.

Output

For each query of the first type output a single integer  — the Mex of{c₀, c₁, ..., c_10⁹}.

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.

注意离散化 ，更换点也要加入数组里面进行离散化

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = 2e5 + ;
 int n, m, tim, L, R, tot, sz, qsz;
 int sum[maxn], cnt[maxn], ans[maxn], now[maxn];
 int a[maxn], b[maxn];
 struct node {
     int l, r, id, t;
     node() {}
     node(int l, int r, int id, int t): l(l), r(r), id(id), t(t) {}
 } qu[maxn];
 struct node1 {
     int pos, x, y;
     node1() {}
     node1(int pos, int x, int y): pos(pos), x(x), y(y) {}
 } c[maxn];
 int cmp(node a, node b) {
     if (a.l / sz == b.l / sz) {
         if (a.r / sz == b.r / sz) return a.t < b.t;
         return a.r < b.r;
     }
     return a.l < b.l;
 }
 void add(int val) {
     cnt[sum[val]]--;
     sum[val]++;
     cnt[sum[val]]++;
 }
 void del(int val) {
     cnt[sum[val]]--;
     sum[val]--;
     cnt[sum[val]]++;
 }
 void change(int pos, int x) {
     if (L <= pos && pos <= R) {
         del(now[pos]);
         add(x);
     }
     now[pos] = x;
 }
 int main() {
     scanf("%d%d", &n, &m);
     sz = (int)pow(n, 0.66666667);
     for (int i =  ; i <= n ; i++) {
         scanf("%d", &a[i]);
         b[++tot] = a[i];
         now[i] = a[i];
     }
     qsz = tim = ;
     for (int i =  ; i <= m ; i++) {
         int op;
         scanf("%d", &op);
         if (op == ) {
             int l, r;
             scanf("%d%d", &l, &r);
             qu[++qsz] = node(l, r, qsz, tim);
         } else {
             int pos, x;
             scanf("%d%d", &pos, &x);
             b[++tot] = x;
             c[++tim] = node1(pos, x, now[pos]);
             now[pos] = x;
         }
     }
     sort(qu + , qu + qsz + , cmp);
     sort(b + , b + tot + );
     tot = unique(b + , b + tot + ) - b;
     for (int i =  ; i <= n ; i++)
         now[i] = lower_bound(b + , b + tot + , a[i]) - b;
     for (int i =  ; i <= tim ; i++) {
         c[i].x = lower_bound(b + , b + tot + , c[i].x) - b;
         c[i].y = lower_bound(b + , b + tot + , c[i].y) - b;
     }
     tim = ;
     for (int i =  ; i <= qsz ; i++) {
         while(L > qu[i].l) add(now[--L]);
         while(R < qu[i].r) add(now[++R]);
         while(L < qu[i].l) del(now[L++]);
         while(R > qu[i].r) del(now[R--]);
         while(tim < qu[i].t) tim++, change(c[tim].pos, c[tim].x);
         while(tim > qu[i].t) change(c[tim].pos, c[tim].y), tim--;
         int mex = ;
         while(cnt[mex] > ) mex++;
         ans[qu[i].id] = mex;
     }
     for (int i =  ; i <= qsz ; i++ )
         printf("%d\n", ans[i]);
     return ;
 }