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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef int64_t ll;

int main(){
  ll n,x=0;
  cin>>n;
  ll a[n];
  for(int i=0;i<n;i++){
    cin>>a[i];
    if(i>1)x^=a[i];
  }
  ll s=a[0]+a[1];
  if(s<x || (s-x)%2!=0){cout<<-1<<endl;return 0;}
  ll d=(s-x)/2;
  if(d>a[0]){cout<<-1<<endl;return 0;}
  for(ll i=0;i<42;i++){
    if(((d>>i)&1)==1 && ((x>>i)&1)==1){cout<<-1<<endl;return 0;}
  }
  ll ans=d;
  for(ll i=41;i>=0;i--){
    if(((x>>i)&1)==1)if(ans+((ll)1<<i)<=a[0])ans+=((ll)1<<i);
  }
  cout<<(ans==0?-1:a[0]-ans)<<endl;
}