Problem: Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself. 

To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed. 

We remind that bracket sequence s is called correct if: 

  * s is empty; 
  * s is equal to "(t)", where t is correct bracket sequence; 
  * s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences. 



For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.

Input

First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.

Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".

Output

Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".

Examples

Input


2
)(


Output


Yes


Input


3
(()


Output


No


Input


2
()


Output


Yes


Input


10
)))))(((((


Output


No

Note

In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.

In the second example, there is no way to move at most one bracket so that the sequence becomes correct.

In the third example, the sequence is already correct and there's no need to move brackets.

Solution (in PYTHON):

Solution:
from __future__ import division, print_function
def main():
    n=int(input())
    stack=[]
    s=input()
    flag=0
    count=0
    for c in s:
        if c=="(":
            stack.append(")")
            count+=1
        else :
            if stack and stack[-1]==c:
                stack.pop()
            elif flag==0:
                stack.append("x")
                flag=1
            else :
                print("No")
                sys.exit()
    if len(stack)==0 or (flag==1 and len(stack)==2):
        print("Yes")
    else :
        print("No")
    
 
 
######## 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
 
if __name__== "__main__":
  main()