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>
/* Uncomment only if you are not using GNU compiler->
*/
#include <map>
#include <algorithm> 
#include <vector> 
#include <string.h> 
#include <math.h> 
#include <stdio.h> 
#include <stdlib.h> 
#include <chrono> 
#include <thread> 
#include <iomanip>
#include <typeinfo>
/**/ 
//define
#define rep(i,n)    for(int i=0;i<n;i++)
#define repu(i,k,n) for(int i=k;i<=n;i++)
#define repd(i,k,n) for(int i=k;i>=n;i--)
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define all(v)  v.begin(), v.end()

//typedef
typedef long long int ll;
typedef long double ld;
typedef std::pair<int, int> pii;
typedef std::vector<int> vi;
typedef std::vector< pii > vii;
typedef std::vector< ll > vll;
typedef std::vector< std::pair < pii, int > > vpp;
 
const long long MOD = 1000000007;
const long long inf = 1e18;
 
using namespace std;

ll modpow(ll x,ll n){
    ll res=1;
    while(n>0){
        if(n&1) res=res*x%MOD;
        x=x*x%MOD;
        n>>=1;
    }
    return res;
}

ll power(ll x, ll n)
{
    ll res=1;
    while(n>0){
        if(n&1) res=res*x;
        x=x*x;
        n>>=1;
    }
    return res;
}

bool custom_sort(int &x, int &y)
{
    if(x<y)
        return 1;
    else
        return 0;
}

void dragnoid99()
{
    int n;
    cin>>n;
    vll a(n);
    ll xorr=0;
    rep(i,n)
    {
        cin>>a[i];
    }
    repu(i,2,n-1)
    {
        xorr^=a[i];
    }
    if(xorr==0)
    {
        cout<<-1<<endl;
        return;
        if((a[0]+a[1])%2!=0 || a[1]>a[0])
        {
            cout<<-1<<endl;
            return;
        }
        cout<<a[0]-(a[0]+a[1])/2<<endl;
        return;
    }
    ll s=a[0]+a[1];
    if((s-xorr)%2==1)
    {
        cout<<-1<<endl;
        return;
    }
    ll x=xorr;
    ll d=(s-x)/2;
    if((d&x))
    {
        cout<<-1<<endl;
        return;
    }
    if(d>a[0])
    {
        cout<<-1<<endl;
        return;
    }
    ll b=0, c=x;
    // cout<<d<<endl;
    // cout<<x<<endl;
    // cout<<(((a[0]-1)&x)==(a[0]-1))<<endl;
    // cout<<(b^d)<<" "<<a[0]<<" "<<x<<endl; 
    repd(i,60,0)
    {
        if(((1LL<<i)&x) && ((b+(1LL<<i))^d)<=a[0])
        {
            b+=(1LL<<i);
        }
    }
    // cout<<b<<endl;
    // cout<<(b^c^x)<<" "<<(b+c==s)<<endl;
    cout<<a[0]-(b^d)<<endl;
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    int t = 1, tt=1;
    // cin>>t;
    while(tt<=t)
    {
        // cout<<"Case #"<<tt<<": ";
        dragnoid99();
        tt++;
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}