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#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef vector<int> VI;

const int INF = 1000000009;
const LL LINF = 1000000000000000009LL;

#define FOR(i, b, e) for (int i = b; i <= e; ++i)
#define FORD(i, b, e) for (int i = b; i >= e; --i)
#define REP(i, n) FOR (i, 0, n - 1)
#define REV(i, n) FORD (i, n - 1, 0)
#define PB push_back
#define PP pop_back
#define MP make_pair
#define MT make_tuple
#define ST first
#define ND second
#define SZ(c) (int)(c).size()
#define ALL(c) (c).begin(), (c).end()
#define UNIQ(c) (c).erase(unique(ALL(c)), (c).end());

// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/hash_policy.hpp>
// using namespace __gnu_pbds;
// typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; /*find_by_order() order_of_key()*/

template <typename T, typename S>
ostream &operator<<(ostream &out, const pair<T, S> &v)
{
    return (out << "(" << v.ST << ", " << v.ND << ")");
}

#define DEBUG(s) s
#ifdef local_comp
#define PRINT(V) cout << "#" << #V << ": " << V << endl;
#else
#define PRINT(v)
#endif

const int N = 500005;

multiset<LL> vals;

LL tab[N];
int n, q;

int solve(LL b, LL e)
{
    // cout << "(" << b << " " << e << ") : ";
    bool rmc = true;
    int res = 0;
    vector<LL> removed;

    while (b < e)
    {
        auto it = vals.lower_bound(b);
        if (it == vals.begin())
        {
            rmc = false;
            break;
        }
        it--;
        if (it != vals.end())
        {
            b += *it;
            // cout << *it << " ";
            removed.PB(*it);
            vals.erase(it);
            res++;
        }
        else
        {
            rmc = false;
            break;
        }
    }
    // cout << "\n";
    for (auto x : removed)
        vals.insert(x);

    if (!rmc)
        return -1;
    return res;
}

int main()
{
#ifndef local_comp
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif

    cin >> n;
    FOR (i, 1, n)
    {
        cin >> tab[i];
        vals.insert(tab[i]);
    }
    cin >> q;
    FOR (i, 1, q)
    {
        int type;
        LL x, y;
        cin >> type >> x;

        if (type == 1)
            cin >> y;

        switch (type)
        {
        case 1:
            cout << solve(x, y) << "\n";
            break;
        case 2:
            vals.insert(x);
            break;
        case 3:
            vals.erase(vals.find(x));
            break;
        }
    }

    return 0;
}