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 Rust (1.42.0) (judge verdict: Time Limit Exceeded)

fn main() {
    let _skip: String = read();
    let a: Vec<i64> = read_vec();

    println!("{}", solve(&a));
}

fn read<T: std::str::FromStr>() -> T {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).ok();
    buf.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|s| s.parse().ok().unwrap())
        .collect()
}

fn solve(a: &[i64]) -> i64 {
    // 石取りゲーム。
    // N個の山のそれぞれの石の数 A(1),A(2),...,A(N) の排他的論理和が0であれば
    // 後手必勝である。

    let x = xor(&a[2..]);

    (0..a[0])
        .find(|i| x ^ (a[0] - i) ^ (a[1] + i) == 0)
        .unwrap_or(-1)
}

fn xor(a: &[i64]) -> i64 {
    a.iter().fold(0, |acc, i| acc ^ i)
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_xor_1() {
        assert_eq!(0, xor(&[0]));
    }

    #[test]
    fn test_xor_2() {
        assert_eq!(1, xor(&[0, 1]));
    }

    #[test]
    fn test_xor_3() {
        assert_eq!(0b11111, xor(&[1, 2, 4, 8, 16]));
    }

    #[test]
    fn test_xor_4() {
        assert_eq!(0b10101, xor(&[1, 2, 4, 8, 16, 10]));
    }

    #[test]
    fn test_solve_1() {
        assert_eq!(1, solve(&[5, 3]));
    }

    #[test]
    fn test_solve_2() {
        assert_eq!(-1, solve(&[3, 5]));
    }

    #[test]
    fn test_solve_3() {
        assert_eq!(-1, solve(&[1, 1, 2]));
    }

    #[test]
    fn test_solve_4() {
        assert_eq!(3, solve(&[10, 9, 8, 7, 6, 5, 4, 3]));
    }

    #[test]
    fn test_solve_5() {
        assert_eq!(1, solve(&[4294967297, 8589934593, 12884901890]));
    }
}