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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ll long long int
#define ld long double
#define ordered_set tree < ll ,  null_type ,  less<ll> ,  rb_tree_tag ,  tree_order_statistics_node_update >
//added two more functions to set
//(1)*(set.find_by_order(k))[kth element in the sorted set] 
//(2)set.order_of_key(k)[count of elements strictly less than k]

typedef vector< int > vi;
typedef vector<long long> lvi;
typedef vector< vector<int> > vvi;
typedef vector< vector<long long> > lvvi;
typedef pair< int,int > ii;
typedef pair< long,long > lii;
typedef vector<pair<int,int>> vii;
typedef vector<pair<long,long>> lvii;
typedef vector<vector<pair<int,int>>> vvii;
typedef vector<vector<pair<long,long>>> lvvii;

#define pb push_back
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(auto i = (c).begin(); i != (c).end(); i++)//remember i is an iterator
#define tc(t) int (t); cin>>(t);while((t)--)
#define ff first
#define ss second
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " is " << a << endl;
	err(++it, args...);
}

//-----------------------------------------------------------------------------------------------------------------------------------------------------------------
#define mod 1000000007
//-----------------------------------------------------------------------------------------------------------------------------------------------------------------

long long mod_sum() { return 0LL; }
template<typename T, typename... Args>
T mod_sum(T a, Args... args) { return ((a + mod_sum(args...))%mod + mod)%mod; }

long long mod_product() { return 1LL; }
template<typename T, typename... Args>
T mod_product(T a, Args... args) { return (a*mod_product(args...))%mod; }
//-----------------------------------------------------------------------------------------------------------------------------------------------------------------


signed main()
{
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);	
	ll n;
	cin>>n;
	ll xrs=0;
	ll a,b; cin>>a>>b;
	for (int i = 0; i < n-2; ++i)
	{
		ll x;
		cin>>x;
		xrs^=x;
	}
	ll sum=a+b;
	ll ands=(sum-xrs)/2;
	if(ands < 0)
	{
		cout<<-1<<endl;
		return 0;
	}
	ll a1,b1;
	a1=b1=ands;
	ll rem=sum - 2*ands;
	if(rem&ands)
	{
		cout<<"-1"<<endl;
		return 0;
	}
	for(int i=50;i>=0;i--)
	{
		if((1ll<<i)&rem)
		{
			if((a1 + (1LL<<i))<=a) a1+=(1LL<<i);
		}
	}
	if(a1>0)
	{
		cout<<a-a1<<endl;
	}
	else cout<<"-1"<<endl;
	return 0;
}