#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <algorithm>

using namespace std;

long long a[300];

int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%lld", &a[i]);
	long long x = 0;
	for (int i = 2; i < n; i++)
		x ^= a[i];
	long long t = a[0] + a[1] - x;
	if (t & 1) {
		puts("-1");
		return 0;
	}
	t /= 2;
	if (t > a[0]) {
		puts("-1");
		return 0;
	}
	for (int i = 39; i >= 0; i--) {
		if (x & (1LL << i)) {
			if (t & (1LL << i)) {
				puts("-1");
				return 0;
			}
			if (t + (1LL << i) <= a[0]) {
				t += (1LL << i);
			}
		}
	}
	if (t == 0) {
		puts("-1");
		return 0;
	}
	printf("%lld\n", a[0] - t);
	return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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.

Examples

Input

2
5 3


Output

1


Input

2
3 5


Output

-1


Input

3
1 1 2


Output

-1


Input

8
10 9 8 7 6 5 4 3


Output

3


Input

3
4294967297 8589934593 12884901890


Output

1

### Response

```cpp
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <algorithm>

using namespace std;

long long a[300];

int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%lld", &a[i]);
	long long x = 0;
	for (int i = 2; i < n; i++)
		x ^= a[i];
	long long t = a[0] + a[1] - x;
	if (t & 1) {
		puts("-1");
		return 0;
	}
	t /= 2;
	if (t > a[0]) {
		puts("-1");
		return 0;
	}
	for (int i = 39; i >= 0; i--) {
		if (x & (1LL << i)) {
			if (t & (1LL << i)) {
				puts("-1");
				return 0;
			}
			if (t + (1LL << i) <= a[0]) {
				t += (1LL << i);
			}
		}
	}
	if (t == 0) {
		puts("-1");
		return 0;
	}
	printf("%lld\n", a[0] - t);
	return 0;
}

```