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>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define ALL(v) (v).begin(), (v).end()
using ll = long long;
constexpr int INF = 1e9;
constexpr long long LINF = 1e18;
constexpr long long MOD = 1e9 + 7;

signed main() {
    int n;
    cin >> n;
    ll a[n];
    rep(i, n) {
        cin >> a[i];
    }
    ll x = 0;
    for (int i = 2; i < n; i++) {
        x ^= a[i];
    }
    int d[40];
    fill(d, d + 40, -1);
    ll b = a[0] + a[1];
    for (int i = 0; i <= 40; i++) {
        if (x & 1LL << i) {
            b -= 1LL << i;
            d[i] = 0;
        }
    }
    ll c = 0;
    for (int i = 40; i >= 0; i--) {
        if (!(x & 1LL << i) && b >= 1LL << (i + 1)) {
            b -= 1LL << (i + 1);
            c += 1LL << i;
            d[i] = 1;
        }
    }
    ll ans = c;
    c = a[0] - c;
    for (int i = 40; i >= 0; i--) {
        if (d[i] == 0 && c >= 1LL << i) {
            ans += 1LL << i;
            c -= 1LL << i;
        }
    }
    ans = a[0] - ans;
    if (0 <= ans && ans < a[0])
        cout << ans << endl;
    else
        cout << -1 << endl;
    return 0;
}