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: Accepted)

#include <bits/stdc++.h>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT_ON(bit, i) (((bit >> i) & 1) > 0)

typedef long long LL;
template<typename T> std::vector<T> make_v(size_t a){return std::vector<T>(a);}
template<typename T,typename... Ts> auto make_v(size_t a, Ts... ts){ return std::vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));}    // C++14
template<typename T,typename V> typename std::enable_if<std::is_class<T>::value==0>::type fill_v(T &t,const V &v){t=v;}
template<typename T,typename V> typename std::enable_if<std::is_class<T>::value!=0>::type fill_v(T &t,const V &v){for(auto &e:t) fill_v(e,v);}
template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline void print_vec(const std::vector<T> &v) { for (int i = 0; i < v.size(); ++i) {  if (i != 0) {std::cout << " ";} std::cout << v[i];} std::cout << "\n"; }
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }
template<typename T> T &chmin(T &a, const T &b) { return a = std::min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = std::max(a, b); }

template<class T> inline std::vector<T> unique(std::vector<T> v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    return v;
}
// 初項s, 交差dのn個の数列の和
long long sum_of_arithmetic_progression(long long s, long long d, long long n) {
    return n * (2 * s + (n - 1) * d) / 2;
}

const int INF = 1 << 30;
const double EPS = 1e-9;
const long double PI = acos(-1.0);
const std::vector<int> dy4 = {0, -1, 0, 1}, dx4 = {1, 0, -1, 0};
const std::vector<int> dy8 = { 0, -1, 0, 1, 1, -1, -1, 1 }, dx8 = { 1, 0, -1, 0, 1, 1, -1, -1 };


template<typename T>class DynamicMedian {
public:
    unsigned int num = 0;
    std::priority_queue<T> min_heap;                                   // 中央値より小さい値を管理
    std::priority_queue<T, std::vector<T>, std::greater<T>> max_heap;  // 中央値より大きい値を管理
    T total_minimum = 0;
    T total_maximum = 0;

    // xを挿入O(log n)
    void insert(T x) {
        if (num % 2 == 0) {
            this->max_heap.push(x);
            this->total_maximum += x;
        } else {
            this->min_heap.push(x);
            this->total_minimum += x;
        }
        balance();
        this->num++;
    }

    // 中央値を見つけるO(1)
    // 存在しない場合は0
    T find_median() {
        // 要素が存在しない
        if (this->num == 0) {
            return 0;
        }

        // サイズが偶数のときは小さい方を中央値とする
        if (num % 2 == 0) {
            return min_heap.top();
        } else {
            return max_heap.top();
        }
    }

    T absolute() {
        T x = this->find_median();
        T mini = std::abs(x * (T)this->min_heap.size() - this->total_minimum);
        T maxi = std::abs(x * (T)this->max_heap.size() - this->total_maximum);

        return mini + maxi;
    }


private:
    void balance() {
        if (min_heap.empty() or max_heap.empty()) {
            return;
        }

        // swap
        if (min_heap.top() > max_heap.top()) {
            T tmp;

            tmp = max_heap.top(); max_heap.pop();
            min_heap.push(tmp);
            this->total_maximum -= tmp;
            this->total_minimum += tmp;

            tmp = min_heap.top(); min_heap.pop();
            max_heap.push(tmp);
            this->total_maximum += tmp;
            this->total_minimum -= tmp;
        }
    }
};

using namespace std;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    DynamicMedian<LL> dm;

    LL Q;
    cin >> Q;

    auto v = make_v<LL>(0);
    LL total_b = 0;

    FOR(i, 0, Q) {
        LL q, a, b;
        cin >> q;

        if (q == 1) {
            cin >> a >> b;
            dm.insert(a);
            total_b += b;
        }
        else {
            LL x = dm.find_median();
            LL ab = dm.absolute();
            cout << x << " " << ab + total_b << endl;
        }
    }

    return 0;
}