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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int dig=52;
inline ll read(){
    ll x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        x=(x<<3)+(x<<1)+c-'0';
        c=getchar();
    }
    return x*f;
}
int n;
ll x,a1,a2;
int main(){
	n=read();
	a1=read(),a2=read();
	for(int i=3;i<=n;++i){
		x^=read();
	}
	ll tmp=a1+a2-x;
	if(tmp&1){
		printf("-1\n");
		return 0;
	}
	tmp>>=1;
	if(a1<=tmp){
		printf("-1\n");
		return 0;
	}
	for(int i=dig;i>=0;--i){
		if((x>>i)&1&&(tmp>>i)&1){
			printf("-1\n");
			return 0;
		}
	}
	ll ans=a1-tmp;
	for(int i=dig;i>=0;--i){
		if((x>>i)&1&&ans>=(1LL<<i)){
			ans-=1LL<<i;
		}
	}
	cout<<(ans==a1?-1:ans)<<endl;
	return 0;
}