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 C (GCC 9.2.1) (judge verdict: Runtime Error)

#include <stdio.h>

int main()
{
	int i, N;
	long long A[301], xor = 0;
	scanf("%d", &N);
	for (i = 1; i <= N; i++) {
		scanf("%lld", &(A[i]));
		xor ^= A[i];
	}
	if (xor == 0) {
		printf("0\n");
		fflush(stdout);
		return 0;
	} else xor ^= A[1] ^ A[2];
	
	long long bit[41], x = A[1] + A[2], y = x - xor;
	for (i = 1, bit[0] = 1; i <= 40; i++) bit[i] = bit[i-1] << 1;
	if (y < 0 || x < y || y % 2 == 1 || y / 2 > A[1]) {
		printf("-1\n");
		fflush(stdout);
		return 0;
	} else y >>= 1;
	
	int flag[41];
	long long a = y, b = y, c = y << 1;
	for (i = 0; i <= 40; i++) flag[i] = y & bit[i]; 
	for (i = 0; i <= 40; i++) {
		if ((x & bit[i]) == (c & bit[i])) continue;
		else if (flag[i] == 1) break;
		flag[i] = -1;
		c += bit[i];
	}
	if (c != x) {
		printf("-1\n");
		fflush(stdout);
		return 0;
	}
	
	for (i = 40; i >= 0; i--) if (flag[i] == -1 && a + bit[i] <= A[1]) a += bit[i];
	if (a > 0) printf("%lld\n", A[1] - a);
	else {
		printf("-1\n");
		return -1;
	}
	fflush(stdout);
	return 0;
}