#include <bits/stdc++.h>
using namespace std;
//为什么这个平衡树也可以用树状数组来做呢？
const int N = 1e5 + 10;

int n, tot, sz;
int op[N], a[N], t[N], c[N];

int pos(int x) {
    return lower_bound(t + 1, t + 1 + sz, x) - t;
}
void add(int x, int k) {
    for (x; x <= sz; x += x & -x) c[x] += k;
}
int query(int x) {
    int tmp = 0;
    for (x; x; x -= x & -x) tmp += c[x];
    return tmp;
}
int kth(int x) {
    int rs = 0;
    for (int i = 17; i >= 0; --i) {
        rs += (1 << i);
        if (rs > sz || c[rs] >= x)
            rs -= (1 << i);
        else
            x -= c[rs];
    }
    return t[rs + 1];
}
int main() {
    scanf("%d", &n);
    //树状数组为啥需要离线处理呢
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &op[i], &a[i]);
        if (op[i] != 4) t[++tot] = a[i];
    }
    //离散化
    sort(t + 1, t + 1 + tot);
    sz = unique(t + 1, t + 1 + tot) - t - 1;

    for (int i = 1; i <= n; i++) {
        int x = a[i];
        if (op[i] == 1) add(pos(x), 1);
        if (op[i] == 2) add(pos(x), -1);
        if (op[i] == 3) printf("%d\n", query(pos(x) - 1) + 1);
        if (op[i] == 4) printf("%d\n", kth(x));
        if (op[i] == 5) printf("%d\n", kth(query(pos(x) - 1)));
        if (op[i] == 6) printf("%d\n", kth(query(pos(x)) + 1));
    }
    return 0;
}