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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

N=int(input())
A=list(map(int,input().split()))
X=0
for i in range(2,N):
  X^=A[i]
B=A[0]
Y=1<<42
P=0
Z=0
for i in range(42):
  Y>>=1
  if Y&(X^A[0]^A[1])==0:
    X=X&(~Y)
    A[0],A[1]=A[0]&(~Y),A[1]&(~Y)
    continue
  Z=min(((~Y)&A[0])+1,(Y-((~Y)&A[1])))
  A[0]-=Z
  A[1]+=Z
  A[0],A[1]=A[0]&(~Y),A[1]&(~Y)
  P+=Z
  X=X&(~Y)
if A[0]^A[1]^X==0 and P<B and P>=0:
  print(P)
else:
  print(-1)