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: Accepted)

#[allow(unused_imports)]
use std::cmp::{max, min};
use std::collections::*;
use std::io::{stdin, Read};

#[allow(unused_macros)]
macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
}

#[allow(unused_macros)]
macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
    ($it: ident, ($($t: tt),*)) => {
        ($(parse_val!($it, $t)),*)
    };
    ($it: ident, u1) => {
        $it.next().unwrap().parse::<usize>().unwrap() -1
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse!(it, n: usize, mut a: usize, b: usize);
    let a0 = a;
    let mut x = 0usize;
    for _ in 0..(n - 2) {
        parse!(it, y: usize);
        x ^= y;
    }
    let m = 43;
    let mut s: usize = a + b;
    let mut dp = vec![vec![vec![-1i64; 2]; 2]; m];
    dp[0][0][0] = 0;
    for i in 0..(m - 1) {
        for j in 0..=1 {
            for k in 0..=1 {
                if dp[i][j][k] == -1 {
                    continue;
                }
                let x = x & 1;
                let s = s & 1;
                let a = a & 1;
                for na in 0..=1 {
                    for nb in 0..=1 {
                        if na ^ nb != x {
                            continue;
                        }
                        if (na + nb + j) % 2 != s {
                            continue;
                        }
                        let nk = match (a, na) {
                            (0, 1) => 1,
                            (1, 0) => 0,
                            _ => k,
                        };
                        let ni = i + 1;
                        let nj = ((na + nb + j) / 2) as usize;
                        dp[ni][nj][nk] = max(dp[ni][nj][nk], dp[i][j][k] + (na << i) as i64);
                    }
                }
            }
        }
        x >>= 1;
        s >>= 1;
        a >>= 1;
    }
    let d = dp[m - 1][0][0];
    if d == -1 || d == 0 {
        println!("{}", -1);
    } else {
        println!("{}", a0 as i64 - d);
    }
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}