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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

fun main() {
  val n = readLine()!!.toInt()
  val a = readLine()!!.split(' ').map { it.toLong() }
  val xor = a.drop(2).fold(0L, Long::xor)
  fun solve(a: Long, b: Long, c: Long): Long {
    if (a xor b == c) return 0L
    val bit = java.lang.Long.highestOneBit(c)
    if (bit == 0L) {
      return if (a <= b || (a + b) % 2 != 0L) -1 else {
        a - (a + b) / 2
      }
    }
    if (a >= bit * 2 || b >= bit * 2) return -1
    if (((a xor b) and bit) == bit) return solve(a and (bit - 1), b and (bit - 1), c xor bit)
    if ((a and bit == bit) && (b and bit == bit)) {
      val t = a - bit + 1
      check(a - t < bit) { "$a $t $b $bit $c"}
      if ((b + t) >= bit * 2) return -1
      return t + solve(a - t, (b + t) and (bit - 1), c xor bit)
    }
    val need = bit - b
    if (a < need) return -1
    return need + solve(a - need, (b + need) and (bit - 1), c xor bit)
  }

  val res = solve(a[0], a[1], xor)
  println(if (res < a[0]) res else -1)
}