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 <stdlib.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef vector<double> Vec;
typedef vector<Vec> Mat;
typedef pair<ll,ll> P;
typedef pair<double,ll> Pd;
typedef priority_queue<P,vector<P>,greater<P>> P_queue;
typedef priority_queue<Pd,vector<Pd>,greater<Pd>> Pd_queue;

const ll MOD=998244353;
const ll mod=1000000007;
const ll INF=1e15;
const double DEL=1e-6;
vec dx={1,0,-1,0};
vec dy={0,1,0,-1};

#define REP(i,a,b) for(int i=(int)a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define mp make_pair
#define ALL(a) a.begin(),a.end()
#define SORT(a) sort(ALL(a))
#define U_ERASE(V) V.erase(unique(ALL(V)), V.end());
#define ADD(a,b) a=(a+b)%mod

ll ruijou(ll n, ll k){
    if(!k) return 1;
    ll a=ruijou(n,k/2);
    a=(a*a);
    return k%2 ? (a*n) : a;
}

ll Solve(){
    ll N; cin>>N;
    vec d(N);
    rep(i,N) cin>>d[i];
    ll Nim=0;
    REP(i,2,N) Nim^=d[i];
    ll Def=d[0]+d[1];
    vec a;
    ll K=Nim;
    if(Def<Nim) return -1;
    if((Def-Nim)%2) return -1;
    while(K>0){
        a.pb(K%2);
        K/=2;
    }

    ll X=Def-Nim;
    X/=2;
    ll P=X;
    vec b;
    while(X>0){
        b.pb(X%2);
        X/=2;
    }
    rep(i,b.size()){
        if(i<a.size())if(b[i] && !a[i]) return -1;
    }
    P=d[0]-P;
    if(P<0) return -1;
    rep(i,a.size()){
        if(ruijou(2,i)>P) return P;
        if(a[i]) P-=ruijou(2,i);
    }
    if(P==d[0]) return -1;
    return P;


}

int main(){
    cout<<Solve()<<endl;
}