这道题显然平衡树，splay，treap什么的随便切
然而我不想打，决定水过这道题
把空间开3倍，树状数组维护它前面的树的个数，开个id数组记录位置
找一个数排名直接二分加求前缀和，log^2的搞一搞
把一个数放在顶/低 直接丢在当前顶/低的前后就可以了不然开3倍数组干嘛
c常数小堪比log的平衡树居然还快一些
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(3e5 + 10), INF(2e9), PF(1e5);
IL ll Read(){
char c = '%'; ll x = 0, z = 1;
for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
return x * z;
}
int n, m, t[_], id[_], a[_];
char opt[10];
IL void Add(RG int x, RG int v){ for(; x <= PF * 3; x += x & -x) t[x] += v; }
IL int Query(RG int x){ RG int cnt = 0; for(; x; x -= x & -x) cnt += t[x]; return cnt; }
IL int Find(RG int x){
RG int l = 1, r = PF * 3;
while(l < r){
RG int mid = (l + r) >> 1;
if(Query(mid) >= x) r = mid;
else l = mid + 1;
}
return l;
}
int main(RG int argc, RG char *argv[]){
n = Read(); m = Read();
for(RG int i = 1; i <= n; ++i) a[PF + i] = Read(), Add(PF + i, 1), id[a[PF + i]] = PF + i;
while(m--){
scanf(" %s", opt); RG int x = Read(), y, p, q;
if(opt[0] == 'T'){
p = Find(1);
a[id[x]] = 0; Add(id[x], -1);
id[x] = p - 1;
a[id[x]] = x; Add(id[x], 1);
}
else if(opt[0] == 'B'){
p = Find(n);
a[id[x]] = 0; Add(id[x], -1);
id[x] = p + 1;
a[id[x]] = x; Add(id[x], 1);
}
else if(opt[0] == 'I'){
y = Read(); q = Query(id[x]);
p = Find(q + y); RG int t = a[p];
swap(a[p], a[id[x]]); swap(id[x], id[t]);
}
else if(opt[0] == 'A') printf("%d
", Query(id[x]) - 1);
else printf("%d
", a[Find(x)]);
}
return 0;
}