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;
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define rep(i,j,n) for(int i=(int)(j);i<(int)(n);i++)
#define REP(i,j,n) for(int i=(int)(j);i<=(int)(n);i++)
#define per(i,j,n) for(int i=(int)(j);(int)(n)<=i;i--)
#define ll long long
#define ALL(a) (a).begin(),(a).end()
#define disup(A,key) distance(A.begin(),upper_bound(ALL(A),(ll)(key)))
#define dislow(A,key) distance(A.begin(),lower_bound(ALL(A),(ll)(key)))
#define pb emplace_back
#define mp std::make_pair
//
#define endl "\n"
//using std::endl;
using std::cin;
using std::cout;
using std::vector;
using std::string;
using std::upper_bound;
using std::lower_bound;
using vi=vector<ll>;
using vii=vector<vi>;
using pii=std::pair<ll,ll>;
//
constexpr ll MOD=1e9+7;
//constexpr ll MOD=998244353;
//constexpr ll MOD=10000000;
//constexpr ll MOD=1e5;
constexpr ll MAX=1e7;
constexpr ll INF=(1ll<<62);
template<class T>
class prique :public std::priority_queue<T, std::vector<T>, std::greater<T>> {};
template<typename T>
struct Segment_tree{
    ll N;
    T mem;
    vector<T> node;
    Segment_tree(vector<T> &X,T m):mem(m){
        ll sz=X.size();
        N=1;
        while(N<sz) N*=2;
        node.resize(2*N-1,mem);
        rep(i,0,sz) node[N-1+i]=X[i];
        per(i,N-2,0){
            node[i]=Compare(node[i*2+1],node[i*2+2]);
        }
    }
    T Compare(T &A,T &B){
        return std::min(A,B);
    }
    void update(ll X,T val){
        X+=N-1;
        node[X]=val;
        while(X>0){
            X=(X-1)/2;
            node[X]=Compare(node[X*2+1],node[X*2+2]);
        }
    }
    T Query(ll a,ll b,ll now,ll l,ll r){ //[a,b),[l,r)
        if(r<0) r=N;
        if(r<=a||b<=l) return mem;
        if(a<=l&&r<=b) return node[now];
        auto vl=Query(a,b,now*2+1,l,(l+r)/2),vr=Query(a,b,now*2+2,(l+r)/2,r);
        return Compare(vl,vr);
    }
};
struct Binary_indexed_tree{
    int N;
    vi bit;
    Binary_indexed_tree(int n):N(n){
        bit.resize(N+1,0);
    }
    void add(int x,int a){
        for(x;x<=N;x+=(x&-x)) bit[x]+=a;
    }
    ll sum(int x){
        ll ret=0;
        for(x;x>0;x-=(x&-x)) ret+=bit[x];
        return ret;
    }
    ll lower_bound(ll X){
        if(sum(N)<X) return -1;
        ll ret=0,memo=1,sum=0;
        while(memo*2<=N) memo*=2;
        while(memo>0){
            if(memo+ret<=N&&sum+bit[memo+ret]<X){
                sum+=bit[memo+ret];
                ret+=memo;
            }
            memo/=2;
        }
        return ret+1;
    }
};
struct Union_Find{
    ll N;
    vi par;
    vi siz;
    Union_Find(int n):N(n){
        par.resize(N);
        siz.resize(N,1);
        rep(i,0,N) par[i]=i;
    }
    ll root(ll X){
        if(par[X]==X) return X;
        return par[X]=root(par[X]);
    }
    bool same(ll X,ll Y){
        return root(X)==root(Y);
    }
    void unite(ll X,ll Y){
        X=root(X);
        Y=root(Y);
        if(X==Y) return;
        par[X]=Y;
        siz[Y]+=siz[X];
        siz[X]=0;
    }
    ll size(ll X){
        return siz[root(X)];
    }
};
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
vi fac,finv,inv;
void COMinit() {
    fac.resize(MAX);
    finv.resize(MAX);
    inv.resize(MAX);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
ll COM(ll n,ll r){
    if(n<r||n<0||r<0) return 0;
    return fac[n]*finv[r]%MOD*finv[n-r]%MOD;
}
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; std::swap(a, b);
        u -= t * v; std::swap(u, v);
    }
    u %= m; 
    if (u < 0) u += m;
    return u;
}
int main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    ll N; cin>>N;
    vi A(N);
    rep(i,0,N) cin>>A[i];
    ll sum=0;
    rep(i,2,N) sum^=A[i];
    ll memo=A[0]+A[1];
    if(memo<sum){
        cout<<-1<<endl;
    }
    else{
        ll P=(memo-sum)>>1;
        ll X=P,Y=P;
        if(X>A[0]||(memo-sum)%2!=0){
            cout<<-1<<endl;
            return 0;
        }
        per(i,62,0){
            if((P&(1ll<<i))==0&&(sum&(1ll<<i))!=0){
                if(X+(1ll<<i)<=A[0]) X+=(1ll<<i);
            }
        }
        if(X==0){
            cout<<-1<<endl;
            return 0;
        }
        cout<<A[0]-X<<endl;
    }
}