One day, Yuhao came across a problem about checking if some bracket sequences are correct bracket sequences.

A bracket sequence is any non-empty sequence of opening and closing parentheses. A bracket sequence is called a correct bracket sequence if it's possible to obtain a correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, the sequences "(())()", "()" and "(()(()))" are correct, while the bracket sequences ")(", "(()" and "(()))(" are not correct.

Yuhao found this problem too simple for him so he decided to make the problem harder. You are given many (not necessarily correct) bracket sequences. The task is to connect some of them into ordered pairs so that each bracket sequence occurs in at most one pair and the concatenation of the bracket sequences in each pair is a correct bracket sequence. The goal is to create as many pairs as possible.

This problem unfortunately turned out to be too difficult for Yuhao. Can you help him and solve it?

Input

The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of bracket sequences.

Each of the following n lines contains one bracket sequence — a non-empty string which consists only of characters "(" and ")".

The sum of lengths of all bracket sequences in the input is at most 5 ⋅ 10^5.

Note that a bracket sequence may appear in the input multiple times. In this case, you can use each copy of the sequence separately. Also note that the order in which strings appear in the input doesn't matter.

Output

Print a single integer — the maximum number of pairs which can be made, adhering to the conditions in the statement.

Examples

Input


7
)())
)
((
((
(
)
)


Output


2


Input


4
(
((
(((
(())


Output


0


Input


2
(())
()


Output


1

Note

In the first example, it's optimal to construct two pairs: "(( )())" and "( )".
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l

q=[]       
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j
"""*******************************************************"""
def main():
    n=inin()
    a=[]
    for i in range(n):
        s=sin()
        j=0
        k=0
        for i in s:
            if(i=="("):
                j+=1
            else:
                j-=1
                if(j<0):
                    k=-1
        z=0
        if(j>=0 and k>=0):
            # print(s)
            a.append(j)
        if(j<0):
            j=k=0
            s=s[::-1]
            # print(s,"ch")
            for i in s:
                if(i==")"):
                    j+=1
                else:
                    j-=1
                    if(j<0):
                        k=-1
            if(j>=0 and k>=0):
                # print(s[::-1])
                a.append(-j)
    d={}
    # print(a)
    for i in a:
        if i in d:
            d[i]+=1
        else:
            d[i]=1
    # print(d)
    ans=0
    for i in d:
        if(i<0):
            if abs(i) in d:
                ans+=min(d[i],d[abs(i)])
    if 0 in d:
        ans+=d[0]//2
    print(ans)
            
        


        
######## 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 = lambda: sys.stdin.readline().rstrip('\r\n')
 
# 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

# threading.Thread(target=main).start()
if __name__== "__main__":
  main()