#include <bits/stdc++.h>

using namespace std;
const int N = 10010;
int a[N];
int q, cmd, number, cnt;
long long p;

//解法1：暴力（使用二分优化查询）
//lower_bound(begin,end,number)函数是求begin-end中（排好序）第一个大于等于 number的数
//upper_bound(begin,end,number)函数是求begin-end中（排好序）第一个大于 number的数
int main() {
    //q次询问
    cin >> q;
    while (q--) {
        cin >> cmd; //1-5共5个命令
        switch (cmd) {
            case 1:
                cin >> number;
                cout << lower_bound(a + 1, a + cnt + 1, number) - a << endl;
                break;
            case 2:
                cin >> number;
                cout << a[number] << endl;
                break;
            case 3:
                cin >> number;
                p = lower_bound(a + 1, a + cnt + 1, number) - a;
                if (p == 1) cout << -2147483647 << endl;
                else cout << a[p - 1] << endl;
                break;
            case 4:
                cin >> number;
                p = upper_bound(a + 1, a + cnt + 1, number) - a;
                if (p == cnt + 1) cout << 2147483647 << endl;
                else cout << a[p] << endl;
                break;
            case 5:
                cin >> number;
                a[++cnt] = number;
                sort(a + 1, a + cnt + 1);
                break;
        }
    }
    return 0;
}