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: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;



int main(){
	cin.tie(0)->sync_with_stdio(0);
	int n;
	cin >> n;
	long long x, y, rest = 0;
	cin >> x >> y;
	for(auto i = 0; i < n - 2; ++ i){
		long long x;
		cin >> x;
		rest ^= x;
	}
	if((x ^ y ^ rest) & 1 || x + y < rest){
		cout << "-1\n";
		return 0;
	}
	long long a = x + y - rest >> 1;
	if(a > x){
		cout << "-1\n";
		return 0;
	}
	long long init_x = x;
	while(x && x & y ^ a){
		for(auto bit = 40; bit >= 0; -- bit){
			if(x & y & 1LL << bit ^ a & 1LL << bit){
				long long d = min(x - (x >> bit << bit) + 1, ((y >> bit) + 1 << bit) - y);
				x -= d, y += d;
				break;
			}
		}
	}
	x ? cout << init_x - x << "\n" : cout << "-1\n";
	return 0;
}

/*

*/

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//                                   Coded by Aeren                                   //
//                                                                                    //
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