#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) {
    return (ull)rng() % B;
}
inline double time() {
    return static_cast<long double>(chrono::duration_cast<chrono::nanoseconds>(chrono::steady_clock::now().time_since_epoch()).count()) * 1e-9;
}

// 0-indexed
template<typename T>
struct BIT{
    int n;
    vector<T> bit,ary;
    BIT(int n = 0) : n(n),bit(n+1),ary(n) {}
    T operator[](int k) {
        return ary[k];
    }
    // [0, i)
    T sum(int i) {
        T res = 0;
        for (; i > 0; i -= (i&-i)) {
            res += bit[i];
        }
        return res;
    }
    // [l, r)
    T sum(int l, int r) {
        return sum(r) - sum(l);
    }
    void add(int i, T a) {
        ary[i] += a;
        i++;
        for (; i <= n; i += (i&-i)) {
            bit[i] += a;
        }
    }
    int lower_bound(T k) { // k <= sum(res)
        if (k <= 0) return 0;
        int res = 0, i = 1;
        while ((i << 1) <= n) i <<= 1;
        for (; i ; i >>= 1) {
            if (res+i <= n and bit[res+i] < k) {
                k -= bit[res += i];
            }
        }
        return res;
    }
    // The 2nd UC Stage 9: Qinhuangdao - I
    // 円環状で見たときに bit[i]+bit[i-1]+...+bit[j] >= k となる最近の j と左辺の総和を求める
    // 0 <= k < bit.sum(n) を仮定　雑にlog2つ
    pair<int, T> lower_bound_ex(int j, T k) {
        T prefix = sum(j+1);
        if (prefix < k) {
            k -= prefix;
            int l = 0, r = n;
            T misum = 0;
            while (r-l > 1) {
                int mid = (l+r)/2;
                T s = sum(mid, n);
                if (s >= k) {
                    misum = s;
                    l = mid;
                }
                else {
                    r = mid;
                }
            }
            return {l, prefix+misum};
        }
        else {
            int l = 0, r = j+1;
            T misum = 0;
            while (r-l > 1) {
                int mid = (l+r)/2;
                T s = sum(mid, j+1);
                if (s > k) {
                    misum = s;
                    l = mid;
                }
                else {
                    r = mid;
                }
            }
            return {l, misum};
        }
    }
};

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n,q; cin >> n >> q;
    ll K; cin >> K;
    vector<ll> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    sort(a.rbegin(), a.rend());
    vector<ll> z(n);
    for (int i = 0; i < n; ++i) {
        z[i] = a[i]%K;
    }
    sort(z.begin(), z.end());
    z.erase(unique(z.begin(), z.end()), z.end());
    int m = z.size();

    int sz = 0;
    BIT<int> bit(m);
    auto add = [&](int i) -> void {
        sz += 1;
        bit.add(i, 1);
    };

    vector<int> md(n);
    for (int i = 0; i < n; ++i) {
        md[i] = lower_bound(z.begin(), z.end(), a[i]%K) - z.begin();
        if (a[0]-a[i] < K) {
            add(md[i]);
        }
    }

    // bit[i]+bit[i-1]+bit[i-2]+...が初めてkを超える位置を求めたい
    // ただしiは既にcnt個使われているものとする
    // 最大値との差分を返すように設計したいねえ
    auto findKth = [&](int i, int k, int cnt) -> ll {
        if (bit[i]-cnt >= k) return 0;
        if (sz-cnt < k) return K;

        k += cnt;
        auto p = bit.lower_bound_ex(i, k);
        ll dif = z[i]-z[p.first];
        if (dif < 0) dif += K;
        return dif;
//
//        int f = k-(bit[i]-cnt);
//        if (bit.sum(i) >= f) {
//            int l = 0, r = i;
//            while (r-l > 1) {
//                int mid = (l+r)/2;
//                if (bit.sum(mid, i) >= f) {
//                    l = mid;
//                }
//                else {
//                    r = mid;
//                }
//            }
//            return z[i]-z[l];
//        }
//        else {
//            f -= bit.sum(i);
//
//            int l = 0, r = m;
//            while (r-l > 1) {
//                int mid = (l+r)/2;
//                if (sz - bit.sum(mid) >= f) {
//                    l = mid;
//                }
//                else {
//                    r = mid;
//                }
//            }
//
//            return z[i]+K-z[l];
//        }
    };

    int pos = md[0];
    int cnt = 0;
    ll mx = a[0];
    __int128 nx = 0;

    auto bitsum = [&](int s, int t) -> int {
        if (t <= s) {
            return bit.sum(t, s+1);
        }
        else {
            return sz-bit.sum(s+1,t);
        }
    };

    auto calNext = [&]() -> void {
        if (sz >= n) {
            return;
        }
        else {
            ll dif = z[pos]-z[md[sz]];
            if (dif < 0) dif += K;
            nx = bitsum(pos, md[sz])-cnt;

            ll uo = mx-dif;
            nx += (__int128)(uo-a[sz]-K)/K*(__int128)sz;
        }
    };

    calNext();

    vector<pair<char,ll>> query;
    auto debug = [&]() -> void {
//        cout << "debug" << endl;
//        for (int i = 0; i < query.size(); ++i) {
//            cout << query[i].first << " " << query[i].second << endl;
//        }
//        cout << "A" << endl;
//        for (int i = 0; i < 18; ++i) {
//            cout << a[i] << " ";
//        }
//        cout << endl;
        for (int i = 0; i < m; ++i) {
            cout << bit[i] << " ";
        }
        cout << endl;
        cout << "pos = " << pos << endl;
        cout << "cnt = " << cnt << endl;
        cout << "mx  = " << mx << endl;
        cout << "nx  = " << (ll)nx << endl;
    };

    while (q--) {
        char c; cin >> c;
        ll uouo; cin >> uouo;
        query.push_back({c, uouo});
        if (c == 'A') {
            int x = uouo;
            ll res;
            if (sz < x) {
                res = a[x-1];
            }
            else {
                ll dif = findKth(pos, x, cnt);
                res = mx-dif;
            }
            cout << res << "\n";
        }
        else {
            auto update = [&]() -> void {
                if (pos) {
                    mx -= z[pos]-z[pos-1];
                    pos -= 1;
                    cnt = 0;
                }
                else {
                    mx -= z[0]-z[m-1]+K;
                    pos = m-1;
                    cnt = 0;
                }
            };

            ll t = uouo;
            while (sz < n and t >= nx) {
                t -= nx;
                mx = a[sz]+K;
                int j = md[sz];
                pos = j;
                add(j);
                cnt = bit[j];

                calNext();
            }

            ll u = t/sz;
            mx -= u*K;
            t -= u*sz;
            nx -= t;

            if (bit[pos]-cnt >= t) {
                cnt += t;
            }
            else {
                t -= (bit[pos]-cnt);
                update();
//                debug();

                if (bit.sum(pos+1) <= t) {
                    t -= bit.sum(pos+1);
                    mx -= z[pos]-z.back()+K;
                    pos = m-1;
                    cnt = 0;
                }

                // (r,pos]は全消し可能
                int l = -1, r = pos;
                while (r-l > 1) {
                    int mid = (l+r)/2;
                    if (bit.sum(mid+1,pos+1) <= t) {
                        r = mid;
                    }
                    else {
                        l = mid;
                    }
                }
                t -= bit.sum(r+1,pos+1);
                mx -= z[pos]-z[r];
                pos = r;
                cnt = 0;
                assert(bit[pos] >= t);
                cnt = t;
            }

//            debug();
        }
    }
}