這道題顯然平衡樹，splay，treap什麼的隨便切
然而我不想打，決定水過這道題
把空間開3倍，樹狀陣列維護它前面的樹的個數，開個id陣列記錄位置
找一個數排名直接二分加求字首和，log^2的搞一搞
把一個數放在頂/低 直接丟在當前頂/低的前後就可以了不然開3倍陣列幹嘛
c常數小堪比log的平衡樹居然還快一些

# 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\n", Query(id[x]) - 1);
        else printf("%d\n", a[Find(x)]);
    }
    return 0;
}