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)

use proconio::input;

fn main() {
    input!{
        n: usize,
        a: [u64; n]
    };
    let s = a[2..n].iter().fold(0, |x, y| x^y);
    let mut dp = [[[a[0]; 2]; 2]; 45];
    dp[0][0][0] = 0;
    for i in 0..44 {
        for j in 0..2 {
            for k in 0..2 {
                if dp[i][j][k] == a[0] { continue; }
                let x = a[0] - (j << i) as u64;
                let y = a[1] + (k << i) as u64;
                if ((x^y^s) >> i) % 2 == 1 { continue }
                let x0 = (a[0] >> i) % 2;
                let y0 = (a[1] >> i) % 2;
                for l in 0..2 {
                    if (l << i) as u64 >= x { continue }
                    let nj = if x0 < (j + l) as u64 { 1 } else { 0 };
                    let nk = if y0 + (k + l) as u64 >= 2 { 1 } else { 0 };
                    dp[i+1][nj][nk] = std::cmp::min(dp[i+1][nj][nk], dp[i][j][k] + (l << i) as u64);
                }
            }
        }
    }
    let mut res = a[0];
    for i in 0..2 {
        for j in 0..2 {
            res = std::cmp::min(res, dp[44][i][j]);
        }
    }
    if res < a[0] {
        println!("{}", res);
    } else {
        println!("-1");
    }
}