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;
typedef long long ll; //int:2*10**9
typedef long double ld;
typedef pair<ll,ll> P;
#define REP(i,n) for(ll i = 0; i<(ll)(n); i++)
#define REPS(i,n) for (ll i = 1; i<=(ll)(n); i++)
#define REPD(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=(a);i<=(b);i++)
#define FORD(i,a,b) for(ll i=(a);i>=(b);i--)
#define vec2(name,i,j) vector<vector<ll>> name(i,vector<ll>(j))
#define vec2f(name,i,j,k) vector<vector<ll>> name(i,vector<ll>(j,k))
#define vec3(name,i,j,k) vector<vector<vector<ll>>> name(i,vector<vector<ll>>(j,vector<ll>(k)))
#define vec3f(name,i,j,k,l) vector<vector<vector<ll>>> name(i,vector<vector<ll>>(j,vector<ll>(k,l)))
#define vec4(name,i,j,k,l) vector<vector<vector<vector<ll>>>> name(i,vector<vector<vector<ll>>>(j,vector<vector<ll>>(k,vector<ll>(l))))
#define vec4f(name,i,j,k,l,m) vector<vector<vector<vector<ll>>>> name(i,vector<vector<vector<ll>>>(j,vector<vector<ll>>(k,vector<ll>(l,m))))
#define pb push_back
#define MOD 1000000007
#define MOD2 998244353
#define PI 3.141592653
#define INF 100000000000000 //14
//cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);

int main(){
    ll n; cin >> n;
    vector<ll> a(n);
    REP(i,n) cin >> a[i];
    ll xo = 0;
    FOR(i,2,n-1) xo=xo^a[i];
    ll s = a[0]+a[1];
    if ((s-xo)%2||s<xo||s-xo>2*a[0]){
        cout << -1 << endl;
        return 0;
    }
    ll ans = (s-xo)/2;
    if (ans&xo!=0){
        cout << -1 << endl;
        return 0;
    }
    FORD(i,50,0) if ((xo>>i)%2&&(1ll<<i)+ans<=a[0]) ans=ans+(1ll<<i);
    if (ans==0) cout << -1 << endl;
    else cout << a[0]-ans << endl;
}