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 (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>

int n;
long long a, b, c, ans, a_, b_, ans_;
short ok = 1;
long long min(long long a, long long b) {
  if(a > b) return b;
  else return a;
}

int main() {
  scanf("%d%lld%lld", &n, &a, &b);
  for(int i = 0; i < n-2; i++) {
    long long d;
    scanf("%lld", &d);
    c ^= d;
  }
  a_ = a;
  b_ = b;
  for(int i = 41; i >= 0; i--) {
    if(c>>i&1) {
      if((1LL<<i) > a+b) ok = 0;
      b -= 1LL<<i;
      ans |= 1LL<<i;
      if(b < 0) {
        a += b;
        b = 0;
      }
    }
    else {
      if(a+b >= (1LL<<(i+1))) {
        a -= 1LL<<i;
        b -= 1LL<<i;
        ans |= 1LL<<i;
        ans_ |= 1LL<<i;
        if(b < 0) {
          a += b;
          b = 0;
        }
      }
    }
  }
  for(int i = 41; i >= 0; i--) {
    if((ans>>i&1) && (~ans_>>i&1)) {
      if((ans_|(1LL<<i)) <= a_) {
        ans_ |= 1LL<<i;
        ans &= ~(1LL<<i);
      }
    }
  }
  if(a || b || ans == a_+b_) ok = 0;
  if(ok) printf("%lld\n", ans-b_);
  else printf("-1\n");
}