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 Pascal (FPC 3.0.4) (judge verdict: Accepted)

var
    n : integer;
    a, b, x, val, y, z, k, res: int64;
    label ending;

begin
    read(n);
    read(a, b);
    n := n - 2;
    while n <> 0 do begin
            n := n - 1;
            read(val);
            x := x xor val;
    end;
    y := a + b - x;
    z := y shr 1;
    if (y < 0) or ((y and 1) <> 0) or ((x and z) <> 0) or (a < z) then
        begin
          writeln('-1');
          goto ending;
        end;
    res := z;
    k := 1 shl 40; 
    while k <> 0 do begin
        if (x and k) <> 0 then begin
            if (res or k) <= a then begin
              res := res or k;
            end;
        end;
        k := k shr 1;
    end;
    if res = 0 then writeln('-1') else writeln(a - res);
    ending:
end.