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 err(args...) {}
#ifdef DEBUG
#include "_debug.cpp"
#endif
using namespace std;
using ll = long long;
using ld = long double;
template <typename T> using lim = numeric_limits<T>;
template <typename T> istream& operator>>(istream& is, vector<T>& a) { for(T& x : a) { is >> x; } return is; }
template <typename X, typename Y> istream& operator>>(istream& is, pair<X, Y>& p) { return is >> p.first >> p.second; }
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<ll> a(n);
    cin >> a;
    ll target_xor = accumulate(a.begin() + 2, a.end(), 0ll, bit_xor<ll>());
    ll target_and = (a[0] + a[1] - target_xor) >> 1;
    ll a0 = 0, a1 = 0;
    bool can = true;
    for(int i = 0; i < 64 and can; i++) {
        if(target_and & (1ll << i)) {
            can &= not (target_xor & (1ll << i));
            a0 |= 1ll << i;
            a1 |= 1ll << i;
        }
    }
    for(int i = 63; i >= 0 and can; i--) {
        if(not (target_and & (1ll << i))) {
            if(target_xor & (1ll << i)) {
                if((a0 | (1ll << i)) <= a[0]) {
                    a0 |= (1ll << i);
                } else {
                    a1 |= (1ll << i);
                }
            }
        }
    }
    cout << (can and 0 < a0 and a0 <= a[0] and a1 >= a[1] ? a[0] - a0 : -1) << endl;
    return 0;
}