I found an interesting problem on https://atcoder.jp/contests/abc172/tasks/abc172_f:
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
 - Starting with Aoki, the two players alternately do the following operation:
 - Operation: Choose one pile of stones, and remove one or more stones from it.
 - When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print -1 instead.

-----Constraints-----
 - 2 \leq N \leq 300
 - 1 \leq A_i \leq 10^{12}

-----Input-----
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N

-----Output-----
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print -1 instead.

-----Sample Input-----
2
5 3

-----Sample Output-----
1

Without moving stones, if Aoki first removes 2 stones from the 1-st pile, Takahashi cannot win in any way.
If Takahashi moves 1 stone from the 1-st pile to the 2-nd before the game begins so that both piles have 4 stones, Takahashi can always win by properly choosing his actions.

I tried it in Python, but could not do it. Can you solve it?
def main():
    import sys
    def input(): return sys.stdin.readline().rstrip()
    n = int(input())
    a = list(map(int, input().split()))

    x = 0
    for i in range(2, n):
        x ^= a[i]
    d = a[0]+a[1]-x
    if d%2 == 1 or d < 0:
        print(-1)
        return
    d >>= 1
    if d&x != 0 or d > a[0]:
        print(-1)
        return
    k = x.bit_length()
    tmp = d
    # d^tmp はd&x=0からd|tmpと一緒
    for i in range(k, -1, -1):
        if (x >> i) & 1:
            if tmp|1<<i <= a[0]:
                tmp |= 1<<i
    if 0 < tmp <= a[0]:
        print(a[0]-tmp)
    else:
        print(-1)



def __starting_point():
    main()
__starting_point()