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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
import itertools
sys.setrecursionlimit(10**5)
stdin = sys.stdin
bisect_left = bisect.bisect_left
bisect_right = bisect.bisect_right
def LI(): return list(map(int, stdin.readline().split()))
def LF(): return list(map(float, stdin.readline().split()))
def LI_(): return list(map(lambda x: int(x)-1, stdin.readline().split()))
def II(): return int(stdin.readline())
def IF(): return float(stdin.readline())
def LS(): return list(map(list, stdin.readline().split()))
def S(): return list(stdin.readline().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')

#A
def A():
    a, b = LI()
    if a <= 5:
        print(0)
    elif 6 <= a <= 12:
        print(b // 2)
    else:
        print(b)
    return

#B
def B():
    r, d, x = LI()
    for i in range(10):
        print(x * r - d)
        x = x * r - d
    return

#C
def C():
    n, m = LI()
    imo = [0 for i in range(n+1)]
    for _ in range(m):
        l, r = LI_()
        imo[l] += 1
        imo[r + 1] -= 1
    for i in range(1,n):
        imo[i] += imo[i - 1]
    ans = 0
    #print(imo)
    for i in imo:
        if i == m:
            ans += 1
    print(ans)
    return

#D
def D():
    n, m = LI()
    a = LI()
    bc = LIR(m)
    bc.sort(key=lambda x: x[1], reverse=True)
    a.sort()
    index = 0
    suma = 0
    for b, c in bc:
        if index + b >= n:
            if a[n - 1] <= c:
                suma += (n - index) * c
            else: 
                br = bisect_left(a, c)
                if br > index:
                    suma += (br - index) * c
                    index = br
                    break
                else:
                    break
            index = n
            break
        if a[index + b - 1] < c:
            suma += b * c
            index += b
        else:
            if a[index] < c:
                bl = bisect_left(a, c)
                suma += (bl - index) * c
                index = bl
            else:
                break
    suma += sum(a[index:])
    print(suma)

    return

#E
def E():
    n, m, k = LI()
    def cmb(n, r, mod):
        if ( r<0 or r>n ):
            return 0
        r = min(r, n-r)
        return g1[n] * g2[r] * g2[n-r] % mod

    g1 = [1, 1] 
    g2 = [1, 1]
    inverse = [0, 1] 

    for i in range( 2, n*m + 1 ):
        g1.append( ( g1[-1] * i ) % mod )
        inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
        g2.append((g2[-1] * inverse[-1]) % mod)
        
    ans = 0
    for d in range(1,n+1):
        ans += d*cmb(n * m - 2, k - 2, mod) * (n - d) * m ** 2 
    for d in range(1,m+1):
        ans += d*cmb(n * m - 2, k - 2, mod) * (m - d) * n ** 2
    print(ans%mod)
    

    
    return

#F
def F():
    q = II()
    small = []
    big = []
    len_small = 0
    len_big = 0
    ans = 0
    ds = defaultdict(int)
    db = defaultdict(int)
    for i in range(q):
        query = LI()

        if query[0] == 1:
            a, b = query[1:3]
            ans += b

            if len_big != len_small:
                small_num = -heappop(small)
                ds[small_num] -= 1

                if small_num <= a:
                    heappush(small, -small_num)
                    heappush(big, a)
                    ans += a - small_num
                    db[a] += 1
                    ds[small_num] += 1

                else:
                    heappush(big, small_num)
                    db[small_num] += 1
                    heappush(small, -a)
                    ds[a] += 1
                    a = -heappop(small)
                    heappush(small, -a)
                    ans -= (len_small - 1 - ds[a] + 1) * (small_num - a)
                    ans += (len_big + 1 - db[a]) * (small_num - a)

                len_big += 1

            else:
                if len_small == 0:
                    heappush(small, -a)

                else:
                    big_num = heappop(big)
                    small_num = -heappop(small)
                    lis = [a, big_num, small_num]
                    lis.sort()
                    heappush(small, -lis[0])
                    ds[lis[0]] += 1
                    heappush(small, -lis[1])
                    ds[lis[1]] += 1
                    heappush(big, lis[2])
                    db[lis[2]] += 1
                    b = -heappop(small)
                    heappush(small, -b)
                    ans += (small_num - b) * (len_big - db[b])
                    ans += (b - small_num) * (len_small - db[b])
                    ans += abs(a - b)

                len_small += 1
        else:
            a = heappop(small)
            print(-a, ans)
            heappush(small, a)
    return


#Solve
if __name__ == '__main__':
    F()