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#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;
}