Problem: When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one. 

Input

The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.

It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.

Output

Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.

Examples

Input


4
ezor


Output


0 


Input


10
nznooeeoer


Output


1 1 0 

Note

In the first example, the correct initial ordering is "zero".

In the second example, the correct initial ordering is "oneonezero".

Solution (in PYTHON3):

Solution:
from math import *
from sys import *
from collections import defaultdict
import os, sys
from io import IOBase, BytesIO
M=10**9+7
def pow(a,b):
    res=1
    while b>0:
        if b&1:
            res*=a
        a*=a
        b>>=1
    return res
def powmod(a,b,m):
    res=1
    while b>0:
        if b&1:
            res=((res*a)%m)
        a*=a
        b>>=1
    return res
def inv(a,m):
    return powmod(a,m-2,m)

def factor(n):
    potentional_p = 3
    itog_list = {}
    if n % 2 == 0:
        itog_list[2] = 0
    while n % 2 == 0:
        n = n // 2
        itog_list[2] += 1
    while n - 1:
        if potentional_p > (n**0.5):
            if n in itog_list:
                itog_list[n] += 1
            else:
                itog_list[n] = 1
            return itog_list
        while n % potentional_p == 0:
            n = n // potentional_p
            if potentional_p in itog_list:
                itog_list[potentional_p] += 1
            else:
                itog_list[potentional_p] = 1
        potentional_p += 2
    return itog_list



def main():
    t=int(input())
    a=input()
    zero=a.count('z')
    one=a.count('n')
    arr=[1]*one+[0]*zero
    print(*arr)








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