You are given two integer sequences, each of length N: a_1, ..., a_N and b_1, ..., b_N.
There are N^2 ways to choose two integers i and j such that 1 \leq i, j \leq N. For each of these N^2 pairs, we will compute a_i + b_j and write it on a sheet of paper.
That is, we will write N^2 integers in total.
Compute the XOR of these N^2 integers.
Definition of XOR
The XOR of integers c_1, c_2, ..., c_m is defined as follows:
 - Let the XOR be X. In the binary representation of X, the digit in the 2^k's place (0 \leq k; k is an integer) is 1 if there are an odd number of integers among c_1, c_2, ...c_m whose binary representation has 1 in the 2^k's place, and 0 if that number is even.
For example, let us compute the XOR of 3 and 5. The binary representation of 3 is 011, and the binary representation of 5 is 101, thus the XOR has the binary representation 110, that is, the XOR is 6.

-----Constraints-----
 - All input values are integers.
 - 1 \leq N \leq 200,000
 - 0 \leq a_i, b_i < 2^{28}

-----Input-----
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
b_1 b_2 ... b_N

-----Output-----
Print the result of the computation.

-----Sample Input-----
2
1 2
3 4

-----Sample Output-----
2

On the sheet, the following four integers will be written: 4(1+3), 5(1+4), 5(2+3) and 6(2+4).

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import bisect\nimport functools\nimport os\nimport sys\n\nsys.setrecursionlimit(10000)\nINF = float(\'inf\')\n\nN = int(input())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\n\ndef debug(fn):\n    if not os.getenv(\'LOCAL\'):\n        return fn\n\n    @functools.wraps(fn)\n    def wrapper(*args, **kwargs):\n        ret = fn(*args, **kwargs)\n        print(\'DEBUG: {}({}) -> \'.format(\n            fn.__name__,\n            \', \'.join(\n                list(map(str, args)) +\n                [\'{}={}\'.format(k, str(v)) for k, v in kwargs.items()]\n            )\n        ), end=\'\')\n        print(ret)\n        return ret\n\n    return wrapper\n\n\n@debug\ndef calc_bit(k):\n    """\n    答えの下から k ビット目を返す\n    :param int k:\n    :return:\n    """\n    # 答えの k ビット目が偶数なら 0、奇数なら 1\n\n    # これ以上の桁は不要\n    mod = 1 << (k + 1)\n    amod = [a & (mod - 1) for a in A]\n    bmod = [b & (mod - 1) for b in B]\n    amod.sort()\n    bmod.sort()\n    ret = 0\n\n    # bmod + amod のうち k ビット目が 1 となる数\n    # => (1 << k) 以上 (2 << k) 未満または (3 << k) 以上 (4 << k) 未満\n    # しゃくとる\n    b01 = bisect.bisect_left(bmod, (1 << k) - amod[0])\n    b10 = bisect.bisect_left(bmod, (2 << k) - amod[0])\n    b11 = bisect.bisect_left(bmod, (3 << k) - amod[0])\n    for a in range(len(amod)):\n        while b01 - 1 >= 0 and bmod[b01 - 1] >= (1 << k) - amod[a]:\n            b01 -= 1\n        while b10 - 1 >= 0 and bmod[b10 - 1] >= (2 << k) - amod[a]:\n            b10 -= 1\n        while b11 - 1 >= 0 and bmod[b11 - 1] >= (3 << k) - amod[a]:\n            b11 -= 1\n        # (1 << k) 以上 (2 << k) 未満\n        ret += b10 - b01\n        # (3 << k) 以上 (4 << k) 未満\n        ret += len(bmod) - b11\n\n    return ret % 2\n\n\nans = 0\nfor i in range(30):\n    ans += calc_bit(i) * (2 ** i)\nprint(ans)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13189