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 <bits/stdc++.h>
#define INF 5000000000000000000
#define ll long long
#define pll pair<ll, ll>
using namespace std;

int main() {
  ll N;
  cin >> N;
  vector<ll> A(N);
  ll either = 0;
  ll sum = 0;
  for (ll i = 0; i < N; ++i) {
    cin >> A.at(i);
    if (i > 1) {
      either ^= A.at(i);
    }
    else {
      sum += A.at(i);
    }
  }
  ll both = sum - either;
  if (both % 2 == 1 || both > A.at(0) * 2 || (either & (both / 2))) {
    cout << -1 << "\n";
    return 0;
  }
  both /= 2;
  ll ans_small = 0;
  for (ll i = 39; i >= 0; --i) {
    ll temp = pow(2, i);
    if (either & temp) {
      if (ans_small + temp + both <= A.at(0)) {
        ans_small += temp;
      }
    }
  }
  if (both + ans_small == 0) {
    cout << -1 << "\n";
    return 0;
  }
  ll ans = A.at(0) - (both + ans_small);
  cout << ans << "\n";
}