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++ (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#include<iostream>
using namespace std;
typedef long long LL;
#define int long long 
LL sum,tXor,ans1,ans2,ans=1e18;
LL a[310];
void dfs(int u,LL s1,LL s2,LL Xor){
	if(ans==1) return ;
	if(s1+s2>sum) return ;
	if(s1>a[1]) return ;
	if(s2+sum-s1<a[2]) return ;
	if(s2-a[2]>=ans||abs(s1%(1ll<(u-1))-a[1]%(1ll<(u-1)))>=ans ) return ;
	if(u&&(s1+s2)%(1ll<<(u-1))!=(sum%(1ll<<(u-1))) ) return ;
	if(Xor==0&&s1+s2==sum&&s1<=a[1]&&a[2]<=s2&&s1){
		ans=min(ans,a[1]-s1);
		return ;
	}
	if(u==45) return ;
	if(Xor&1){
		dfs(u+1,s1+(1ll<<u),s2,Xor/2);
		dfs(u+1,s1,s2+(1ll<<u),Xor/2);
	}
	else {
		dfs(u+1,s1+(1ll<<u),s2+(1ll<<u),Xor/2);
		dfs(u+1,s1,s2,Xor/2);
	}
}
signed main(){
	int n,k=0;
	cin>>n;
	for(int i=1;i<=n;++i) {
		cin>>a[i];
		k=k^a[i];
	}
	if(k==0){
		cout<<"0\n";
		return 0;
	}
	sum=a[1]+a[2];
	ans1=-1,ans2=-1;
	tXor=(k^a[1]^a[2]);
	dfs(0,0,0,tXor);
	if(ans!=1e18){
		cout<<ans<<endl;
	}
	else cout<<"-1"<<endl;
	return 0;
}