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 = '"""\n感想\n- 桁dp,その桁に1立つか調べる\n- 桁に1が立つ=値の範囲が[T,2T), [3T, 4T)...\n- その桁をしるためにはmod 2*Tの値だけ見れば十分\n- 値の範囲の個数をbisectで探る\n- bisectだとpythonだと死ぬ\n  - NlogNはだめ?\n  - 尺取で2*Nにする\n  - でもソートしてるからNlogNじゃ？\n  - 定数倍のギリギリの問題？=>bisect一個ならセーフ, 探索範囲絞ってbisectのコードは本当にギリギリで運次第でAC(同じコードでTLEとAC)\n"""\nN = int(input())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\nans = 0\nfor i in range(29):\n    div_ = pow(2, i+1)\n    A_tmp = sorted([x%div_ for x in A])\n    B_tmp = sorted([x%div_ for x in B])\n    cnt = 0\n    tmp = pow(2, i)\n    idx1 = N\n    idx2 = N\n    idx3 = N\n    for a in A_tmp:\n        while idx2>0 and B_tmp[idx2-1]>=2*tmp-a:\n            idx2 -= 1\n        while idx1>0 and B_tmp[idx1-1]>=tmp-a:\n            idx1 -= 1\n        while idx3>0 and B_tmp[idx3-1]>=3*tmp-a:\n            idx3 -= 1\n        cnt += ((idx2-idx1)+(N-idx3))%2\n    if cnt % 2 == 1:\n        ans += tmp\nprint(ans)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13587