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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define MAX_AR 300

using namespace std;

typedef long long ll;
typedef pair<int, int> ii;

ll a1, a2, tot = 0;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;
    ll t;
    for (int ctr1 = 0; ctr1 < n; ++ctr1) {
        cin >> t;
        if (ctr1 == 0)
            a1 = t;
        else if (ctr1 == 1)
            a2 = t;
        else tot ^= t;
    }

    ll s = a1 + a2;
    ll aib = (s - tot) / 2;
    if (aib < 0 || a1 < aib || (aib & tot)) {
        cout << "-1\n";
    } else {
        ll rez = 0;
        for (int ctr1 = 40; ctr1 >= 0; --ctr1) {
            if (!((1LL << ctr1) & aib)) {
                if (rez + (1LL << ctr1) + aib <= a1 && (tot & (1LL << ctr1)))
                    rez += (1LL << ctr1);
            }
        }
        if ((rez ^ aib) == 0) {
            cout << "-1\n";
        } else {
            cout << a1 - (rez ^ aib) << '\n';
        }
    }

    return 0;
}