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>
#define ll long long
#define dl double
#define mp make_pair
#define pb push_back
using namespace std;

//***********************************************************************

// hash struct for unordered_map
struct hash_pair { 
    template <class T1, class T2> 
    size_t operator()(const pair<T1, T2>& p) const
    { 
        auto hash1 = hash<T1>{}(p.first); 
        auto hash2 = hash<T2>{}(p.second); 
        return hash1 ^ hash2; 
    } 
};

int MAXNUM = 5e3+5;
vector <int>vis(MAXNUM);
vector<int> pr;
void sieve() {
	pr.push_back(2);
	for(ll i=3;i<MAXNUM;i+=2) if(!vis[i]) {
		pr.push_back(i);
		for(ll j=i*i;j<MAXNUM;j+=2*i) vis[j]=true;
	}
}
//************************************************************************

void solve()
{
    ll n;
    cin >> n;
    ll S=0;
    ll X=0;
    ll a1;
    for(int i=0;i<n;i++)
    {
        ll x;
        cin >> x;
        if(i==0)
        a1=x;
        if(i<2)
        S+=x;
        if(i>1)
        X^=x;
    }
    ll D=(S-X)/2;
    if(D<0 || (D&X))
    {
        cout <<-1 << endl;
        return;
    }
    ll ans=D;
    if(D>a1)
    {
        cout <<-1 << endl;
        return;
    }
    for(ll i=42;i>=0;i--)
    {
        if((ll)pow(2,i)&X)
        if(ans+(ll)pow(2,i)<=a1)
        ans+=(ll)pow(2,i);
    }
    if(ans==0)
    cout << -1 << endl;
    else
    cout << a1-ans << endl;
}


int main()
{ 
	ios_base::sync_with_stdio(false); 
    cin.tie(NULL);
	ll t;
	//cin >> t;
	t=1;
	while(t--)
	{
		solve();
	}

	return 0;
}