p03040 AtCoder Beginner Contest 127 - Absolute Minima_13721
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.

Examples

Input

4
1 4 2
2
1 1 -8
2


Output

4 2
1 -3


Input

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


Output

-1000000000 3000000000
import heapq

def main(): 
    smaller = []
    larger = []

    smallsum = 0
    largesum = 0

    const = 0

    q = int(input())
    for _ in range(q):
        inp = input()
        if inp[0] == '1':
            _,a,b = [int(x) for x in inp.split()]
            const += b

            if not smaller:
                smaller.append(-a)
                smallsum += a

            elif a <= -smaller[0]:
                heapq.heappush(smaller, -a)
                smallsum += a

            else:
                heapq.heappush(larger, a)
                largesum += a

            while len(smaller)>len(larger)+1:
                a = -heapq.heappop(smaller)
                smallsum -= a
                heapq.heappush(larger, a)
                largesum += a
            while len(larger)>len(smaller):
                a = heapq.heappop(larger)
                largesum -= a
                heapq.heappush(smaller,-a)
                smallsum += a

        else:
            x = (-smaller[0]) if smaller else (-10**9)
            print x, const + len(smaller)*x - smallsum + largesum - len(larger)*x


######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = sys.stdin.readline

# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'

# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

if __name__== "__main__":
  main()

