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: Wrong Answer)

use itertools::Itertools;
use proconio::input;
use std::cmp::min;

const WIDTH: usize = 40;

fn main() {
    input!(n: usize, orig: [u64; n]);
    let make_seq = |x| (0..WIDTH).map(|i| (x >> i & 1) as usize).collect_vec();
    let a = make_seq(orig[0]);
    let b = make_seq(orig[1]);
    let c = make_seq({
        let mut x = 0;
        for y in orig.iter().skip(2) {
            x ^= y;
        }
        x
    });
    // eprintln!("{:?}", a);
    // eprintln!("{:?}", b);
    // eprintln!("{:?}", c);

    let mut dp = vec![vec![[std::u64::MAX; 2]; 2]; WIDTH + 1];
    dp[0][0][0] = 0;
    for i in 0..WIDTH {
        let x = a[i];
        let y = b[i];
        let z = c[i];
        // eprintln!("(i, x, y, z) = {:?}", (i, x, y, z));
        // eprintln!("dp[i] = {:?}", dp[i]);
        for ((j, k), l) in (0..2)
            .cartesian_product(0..2)
            .cartesian_product(0..2)
            .filter(|&((j, k), _)| x ^ y ^ z ^ j ^ k == 0)
        {
            if dp[i][j][k] != std::u64::MAX {
                let nj = (x < j + l) as usize;
                let nk = (2 <= y + l + k) as usize;
                dp[i + 1][nj][nk] = min(dp[i + 1][nj][nk], dp[i][j][k] + l as u64 * (1 << i));
            }
        }
    }
    let ans = *dp[WIDTH]
        .iter()
        .map(|v| v.iter().min().unwrap())
        .min()
        .unwrap();
    println!("{}", if orig[0] <= ans { -1 } else { ans as i64 });
}