Problem:
Score : 600 points

Problem Statement

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 1

2
5 3

Sample Output 1

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.

Sample Input 2

2
3 5

Sample Output 2

-1

It is not allowed to move stones from the 2-nd pile to the 1-st.

Sample Input 3

3
1 1 2

Sample Output 3

-1

It is not allowed to move all stones from the 1-st pile.

Sample Input 4

8
10 9 8 7 6 5 4 3

Sample Output 4

3

Sample Input 5

3
4294967297 8589934593 12884901890

Sample Output 5

1

Watch out for overflows.

Solution in Python (3.8.2) (judge verdict: Wrong Answer)

import sys


def solve(n, aaa):
    if n == 2:
        h, m = divmod(sum(aaa), 2)
        if m == 1 or h > aaa[0]:
            return -1
        return aaa[0] - h

    x = 0
    for a in aaa[2:]:
        x ^= a
    a0, a1 = aaa[:2]
    s = a0 + a1

    if s & 1 != x & 1:
        return -1

    p, q = 0, 0

    b = 1 << 40
    while b:
        bx = x & b
        if bx == 0:
            # print('bx', s)
            # print(f'{b:040b}')
            # print(f'{p:040b}')
            # print(f'{q:040b}')
            # print(f'{a0:040b}')
            # print(f'{a1:040b}')
            # print()
            if s >= 2 * b:
                if p | b <= a0:
                    p |= b
                    q |= b
                    s -= 2 * b
                else:
                    return -1
        else:
            if s < b:
                return -1
            if p | b <= a0 and q | b > a1:
                p |= b
            else:
                q |= b
            s -= b

        # print(f'{b:040b}')
        # print(f'{p:040b}')
        # print(f'{q:040b}')
        # print(f'{a0:040b}')
        # print(f'{a1:040b}')
        # print()

        b >>= 1

    if p == 0:
        return -1
    assert s == 0
    assert p ^ q == x
    assert a0 - p == q - a1
    return a0 - p


n, *aaa = map(int, sys.stdin.buffer.read().split())
print(solve(n, aaa))