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: Accepted)

#include <iostream>
#include <algorithm>
#include <utility>
#include <vector>
#include <string>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define REP(i, n) for (int i = 0; i < (n); i++)
#define RREP(i, n) for (int i = (n) - 1; i >= 0; i--)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (a); i > (b); i--)
#define ALL(a) (a).begin(), (a).end()

int main() {
    int N;
    cin >> N;
    ll A, B, S, X = 0;
    cin >> A >> B;
    S = A + B;
    REP(i, N - 2) {
        ll tmp; cin >> tmp;
        X ^= tmp;
    }

    ll dp[42][2][2] = {};
    REP(i, 42) REP(j, 2) REP(k, 2) dp[i][j][k] = -1;
    dp[0][0][0] = 0;
    REP(i, 41) {
        int cx = (X >> i) & 1;
        int cs = (S >> i) & 1;
        int ca = (A >> i) & 1;

        REP(j, 2) REP(k, 2) {
            if (dp[i][j][k] == -1) continue;

            REP(na, 2) {
                int nb = cx ^ na;
                int ns = na + nb + j;
                if ((ns & 1) != cs) continue;

                int ni = i + 1, nj, nk;
                if (ns >= 2) nj = 1;
                else nj = 0;

                if (na > ca) nk = 1;
                else if (na == ca) nk = k;
                else nk = 0;

                dp[ni][nj][nk] = max(dp[ni][nj][nk], dp[i][j][k] | ((ll)na << i));
            }
        }
    }

    ll a = dp[41][0][0];
    if (a == -1 || a == 0) cout << -1 << endl;
    else cout << A - a << endl;
    return 0;
}