Problem: Polycarp is a head of a circus troupe. There are n — an even number — artists in the troupe. It is known whether the i-th artist can perform as a clown (if yes, then c_i = 1, otherwise c_i = 0), and whether they can perform as an acrobat (if yes, then a_i = 1, otherwise a_i = 0).

Split the artists into two performances in such a way that: 

  * each artist plays in exactly one performance, 
  * the number of artists in the two performances is equal (i.e. equal to n/2), 
  * the number of artists that can perform as clowns in the first performance is the same as the number of artists that can perform as acrobats in the second performance. 

Input

The first line contains a single integer n (2 ≤ n ≤ 5 000, n is even) — the number of artists in the troupe.

The second line contains n digits c_1 c_2 … c_n, the i-th of which is equal to 1 if the i-th artist can perform as a clown, and 0 otherwise.

The third line contains n digits a_1 a_2 … a_n, the i-th of which is equal to 1, if the i-th artist can perform as an acrobat, and 0 otherwise.

Output

Print n/2 distinct integers — the indices of the artists that should play in the first performance.

If there are multiple answers, print any.

If there is no solution, print a single integer -1.

Examples

Input


4
0011
0101


Output


1 4


Input


6
000000
111111


Output


-1


Input


4
0011
1100


Output


4 3


Input


8
00100101
01111100


Output


1 2 3 6

Note

In the first example, one of the possible divisions into two performances is as follows: in the first performance artists 1 and 4 should take part. Then the number of artists in the first performance who can perform as clowns is equal to 1. And the number of artists in the second performance who can perform as acrobats is 1 as well.

In the second example, the division is not possible.

In the third example, one of the possible divisions is as follows: in the first performance artists 3 and 4 should take part. Then in the first performance there are 2 artists who can perform as clowns. And the number of artists in the second performance who can perform as acrobats is 2 as well.

Solution (in PYTHON3):

Solution:
"""
    Author - Satwik Tiwari .
    13th NOV , 2020  - Friday
"""

#===============================================================================================
#importing some useful libraries.


from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key

# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt

from copy import deepcopy
from collections import deque


from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect

#==============================================================================================
#fast I/O region
BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

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


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

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

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

#===============================================================================================
#some shortcuts

def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
    for pp in range(t):
        solve(pp)
def google(p):
    print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
    y%=(p-1)  #not so sure about this. used when y>p-1. if p is prime.
    res = 1     # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0) :
        return 0
    while (y > 0) :
        if ((y & 1) == 1) : # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1      # y = y/2
        x = (x * x) % p
    return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))



def solve(case):
    n = int(inp())
    c = list(inp())
    b = list(inp())

    a = [0,0,0,0]
    for i in range(n):
        if(c[i] == '0' and b[i] == '0'):
            a[0]+=1
        if(c[i] == '1' and b[i] == '1'):
            a[1]+=1
        if(c[i] == '0' and b[i] == '1'):
            a[3] +=1
        if(c[i] == '1' and b[i] == '0'):
            a[2]+=1

    # print(a)
    for i in range(a[1]+a[2]+1):
        clowns = i
        if(clowns>(n//2)):
            continue
        for j in range(0,i+1):
            both = a[1] - j
            cl = clowns - j

            if(cl>a[2] or both<0):
                continue
            # print(clowns,both)
            if(a[3]+both < clowns or clowns +a[2] - cl>n//2 or both>clowns):
                continue

            ac = (clowns - both)
            if(a[3] - ac+clowns > n//2):
                continue

            # print(ac,a,both,clowns,cl)
            ans = []
            cnt = 0
            ind = 0
            taken = [0]*n

            while(cnt<j):
                if(c[ind] == '1' and b[ind] == '1' and not taken[ind]):
                    cnt+=1
                    ans.append(ind)
                    taken[ind] = 1
                ind+=1
            ind = 0
            while(cnt<clowns):
                if(c[ind] == '1' and b[ind] == '0' and not taken[ind]):
                    cnt+=1
                    ans.append(ind)
                    taken[ind] = 1
                ind+=1
            ind = 0
            while(cnt<clowns + a[3] - ac):
                if(c[ind] == '0' and b[ind] == '1' and not taken[ind]):
                    cnt+=1
                    ans.append(ind)
                    taken[ind] = 1
                ind+=1
            ind = 0
            while(cnt<n//2):
                if(c[ind] == b[ind] and c[ind] == '0' and not taken[ind]):
                    cnt+=1
                    ans.append(ind)
                    taken[ind] = 1
                ind+=1

            print(' '.join(str(ans[ii]+1) for ii in range(n//2)))
            return




        # solved my own problem. its just stupid.
        # the real one is easy version of what i just solved.
        #
        # if(a[3] <= clowns):
        #     ans = []
        #     cnt = 0
        #     ind = 0
        #     taken = [0]*n
        #     while(cnt<clowns and ind<n):
        #         if(c[ind] == '1' and b[ind] == '0' and not taken[ind]):
        #             cnt+=1
        #             ans.append(ind)
        #             taken[ind] = 1
        #         ind+=1
        #     ind = 0
        #     while(cnt<clowns and ind<n):
        #         if(c[ind] == '1' and b[ind] == '1' and not taken[ind]):
        #             cnt+=1
        #             ans.append(ind)
        #             taken[ind] = 1
        #         ind+=1
        #     ind = 0
        #     while(cnt<n//2):
        #         if(c[ind] == b[ind] and not taken[ind]):
        #             ans.append(ind)
        #             taken[ind] = 1
        #             cnt+=1
        #         ind+=1
        #
        #     print(' '.join(str(ans[ii]+1) for ii in range(n//2)))
        #     return
        # else:
        #     rem = a[3] - clowns
        #     if(rem>(n//2) - clowns):
        #         continue
        #     ans = []
        #     cnt = 0
        #     ind = 0
        #     taken = [0]*n
        #     while(cnt<clowns and ind<n):
        #         if(c[ind] == '1' and b[ind] == '0' and not taken[ind]):
        #             cnt+=1
        #             ans.append(ind)
        #             taken[ind] = 1
        #         ind+=1
        #     ind = 0
        #     while(cnt<clowns and ind<n):
        #         if(c[ind] == '1' and b[ind] == '1' and not taken[ind]):
        #             cnt+=1
        #             ans.append(ind)
        #             taken[ind] = 1
        #         ind+=1
        #     ind = 0
        #     while(cnt<clowns+rem):
        #         if(c[ind] == '0'and b[ind] == '1' and not taken[ind]):
        #             cnt+=1
        #             ans.append(ind)
        #             taken[ind] = 1
        #         ind+=1
        #     ind = 0
        #     # print(clowns,rem,cnt)
        #     while(cnt<n//2):
        #         if(c[ind] == b[ind] and not taken[ind]):
        #             cnt+=1
        #             ans.append(ind)
        #             taken[ind] = 1
        #         ind+=1
        #     print(' '.join(str(ans[ii]+1) for ii in range(n//2)))
        #     return

    print(-1)




testcase(1)
# testcase(int(inp()))