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;
using Int = int_fast64_t;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	Int n; cin >> n;
	vector<Int> a(n);
	for(auto &i:a) cin >> i;
	Int k = 0;
	for(size_t i=2; i<a.size(); ++i)
		k ^= a[i];
	Int l = a[0] + a[1];
	if(l - k < 0 || (l - k) % 2 == 1){
		cout << -1 << "\n";
		return 0;
	}
	Int ans = (l - k) / 2;
	for(Int i=1; i<1ll<<41; i<<=1)
		if((ans & i) != 0 && k & i == 0){
			cout << -1 << "\n";
			return 0;
		}
	for(Int i=1ll<<40; i>0; i>>=1)
		if((i & k) != 0 && ans + i <= a[0])
			ans += i;
	cout << (ans == 0 || ans > a[0] ? -1 : a[0] - ans) << "\n";
}