Problem:
Score : 600 points

Problem Statement

There is a function f(x), which is initially a constant function f(x) = 0.

We will ask you to process Q queries in order. There are two kinds of queries, update queries and evaluation queries, as follows:

An update query 1 a b: Given two integers a and b, let g(x) = f(x) + |x - a| + b and replace f(x) with g(x).

An evaluation query 2: Print x that minimizes f(x), and the minimum value of f(x). If there are multiple such values of x, choose the minimum such value.

We can show that the values to be output in an evaluation query are always integers, so we ask you to print those values as integers without decimal points.

Constraints

All values in input are integers.

1 \leq Q \leq 2 \times 10^5

-10^9 \leq a, b \leq 10^9

The first query is an update query.

Input

Input is given from Standard Input in the following format:

Q
Query_1
:
Query_Q

See Sample Input 1 for an example.

Output

For each evaluation query, print a line containing the response, in the order in which the queries are given.

The response to each evaluation query should be the minimum value of x that minimizes f(x), and the minimum value of f(x), in this order, with space in between.

Sample Input 1

4
1 4 2
2
1 1 -8
2

Sample Output 1

4 2
1 -3

In the first evaluation query, f(x) = |x - 4| + 2, which attains the minimum value of 2 at x = 4.

In the second evaluation query, f(x) = |x - 1| + |x - 4| - 6, which attains the minimum value of -3 when 1 \leq x \leq 4. Among the multiple values of x that minimize f(x), we ask you to print the minimum, that is, 1.

Sample Input 2

4
1 -1000000000 1000000000
1 -1000000000 1000000000
1 -1000000000 1000000000
2

Sample Output 2

-1000000000 3000000000

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;

#define REP(i,n) for(long long i = 0; i < (n); i++)
#define FOR(i, m, n) for(long long i = (m);i < (n); ++i)
#define ALL(obj) (obj).begin(),(obj).end()
#define SPEED cin.tie(0);ios::sync_with_stdio(false);

template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQR = priority_queue<T,vector<T>,greater<T>>;

const ll MOD = (ll)1e9 + 7;
const ll MOD2 = 998244353;
const ll HIGHINF = (ll)1e18;
const ll LOWINF = (ll)1e15;
const long double PI = 3.1415926535897932384626433;

template<typename T> vector<T> make_v(size_t N,T init){return vector<T>(N,init);}
template<typename... T> auto make_v(size_t N,T... t){return vector<decltype(make_v(t...))>(N,make_v(t...));}
template <class T> void corner(bool flg, T hoge) {if (flg) {cout << hoge << endl; exit(0);}}
template <class T, class U>ostream &operator<<(ostream &o, const map<T, U>&obj) {o << "{"; for (auto &x : obj) o << " {" << x.first << " : " << x.second << "}" << ","; o << " }"; return o;}
template <class T>ostream &operator<<(ostream &o, const set<T>&obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
template <class T>ostream &operator<<(ostream &o, const multiset<T>&obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
template <class T>ostream &operator<<(ostream &o, const vector<T>&obj) {o << "{"; for (int i = 0; i < (int)obj.size(); ++i)o << (i > 0 ? ", " : "") << obj[i]; o << "}"; return o;}
template <class T, class U>ostream &operator<<(ostream &o, const pair<T, U>&obj) {o << "{" << obj.first << ", " << obj.second << "}"; return o;}
template <template <class tmp>  class T, class U> ostream &operator<<(ostream &o, const T<U> &obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr)o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
void print(void) {cout << endl;}
template <class Head> void print(Head&& head) {cout << head;print();}
template <class Head, class... Tail> void print(Head&& head, Tail&&... tail) {cout << head << " ";print(forward<Tail>(tail)...);}
template <class T> void chmax(T& a, const T b){a=max<T>(a,b);}
template <class T> void chmin(T& a, const T b){a=min<T>(a,b);}
void YN(bool flg) {cout << ((flg) ? "YES" : "NO") << endl;}
void Yn(bool flg) {cout << ((flg) ? "Yes" : "No") << endl;}
void yn(bool flg) {cout << ((flg) ? "yes" : "no") << endl;}

int main() {
    int Q,flg; cin >> Q;
    ll sum,cen,sumL = 0,sumR = 0,L,R; cin >> flg >> cen >> sum;
    PQR<ll> pqR;
    PQ<ll> pqL;
   
    for(int i = 1; i < Q; ++i){
        int q; cin >> q; q--;
        if(!q){
            ll a,b; cin >> a >> b;
            sum += b;
            if(flg){
                if(cen>=a) swap(cen,a);
                L = cen,R = a;
                cen = (cen+a)/2;
            }
            else{
                L = pqL.top();
                pqL.pop();
                sumL -= L;
                R = pqR.top();
                pqR.pop();
                sumR -= R;
                V<ll> tmp;
                tmp.push_back(a);
                tmp.push_back(L);
                tmp.push_back(R);
                sort(ALL(tmp));
                L = tmp[0];
                cen = tmp[1];
                R = tmp[2];
            }
            flg^=1;
            pqL.push(L);
            sumL += L;
            pqR.push(R);
            sumR += R;          
        }
        else{
            ll sL = (ll)pqL.size();
            ll sR = (ll)pqR.size();
            ll ans = sum + (cen*sL) - sumL + sumR - (cen*sR);
           
            ll ok = cen,ng = L-1,md;
            while(ok-ng>1){
                md = (ok+ng)/2;
                ll cnt = sum + (md*sL) - sumL + sumR - (md*sR);
                (cnt<=ans?ok:ng)=md;
            }
            cen = ok;
            
            cout << cen << " " << sum + (cen*sL) - sumL + sumR - (cen*sR) << endl;
        }
    }



    return 0;
}