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)

// 解説のDP
// y + (y ^ xor) = sum, y <= sup を満たすyの最大値を求める
// 下からやっていく
// [i桁目][i桁目まで見たとき未満である][carry] = 最大値
fn solve(sum: usize, xor: usize, sup: usize) -> Option<usize> {
    let mut dp = vec![vec![None; 2]; 2];
    dp[0][1] = Some(0);
    for i in 0..41 {
        let mut next = vec![vec![None; 2]; 2];
        for (cond, dp) in dp.iter().enumerate() {
            for (carry, v) in dp.iter().enumerate().filter(|v| v.1.is_some()) {
                let v = v.unwrap();
                let x = sum >> i & 1;
                let y = xor >> i & 1;
                let z = sup >> i & 1;
                for a in 0..2 {
                    if (a + (y ^ a) + carry) & 1 != x || v + (a << i) > sup {
                        continue;
                    }
                    let carry = (a + (y ^ a) + carry) >> 1;
                    let cond = ((cond == 1 && a <= z) || (a == 0 && z == 1)) as usize;
                    let v = v + (a << i);
                    let p = next[cond][carry].get_or_insert(v);
                    *p = std::cmp::max(*p, v);
                }
            }
        }
        dp = next;
    }
    let mut ans = None;
    for p in dp.into_iter().flatten() {
        p.map(|p| {
            let q = ans.get_or_insert(p);
            *q = std::cmp::min(*q, sup - p);
        });
    }
    ans
}

fn run() {
    proconio::input! {
        n: usize,
        a: [usize; n],
    }
    let xor = a.iter().skip(2).fold(0, |s, a| s ^ *a);
    let sum = a[0] + a[1];
    if let Some(ans) = solve(sum, xor, a[0]) {
        println!("{}", ans);
    } else {
        println!("-1");
    }
}

fn main() {
    run();
}