#include <bits/stdc++.h>
using namespace std;

int main() {
  ios::sync_with_stdio(false);

  int N;
  cin >> N;

  vector<int64_t> A(N);
  for (int i = 0; i < N; ++i) cin >> A[i];

  int64_t P = A[0] + A[1];
  int64_t S = 0; for (int i = 2; i < N; ++i) S ^= A[i];

  if (P - S & 1) {
    cout << -1 << endl;
    return 0;
  }

  int64_t D = P - S >> 1;

  int64_t a = D;
  int64_t b = D;
  for (int i = 40; ~i; --i) if (S >> i & 1) {
    int64_t na = a | (int64_t) 1 << i;
    if (na <= A[0]) a = na;
    else b = b | (int64_t) 1 << i;
  }

  if (a + b == P && (a ^ b) == S && (1 <= a && a <= A[0])) {
    cout << A[0] - a << endl;
  } else {
    cout << -1 << endl;
  }

  return 0;
}


### Prompt

In Cpp, your task is to solve 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 <bits/stdc++.h>
using namespace std;

int main() {
  ios::sync_with_stdio(false);

  int N;
  cin >> N;

  vector<int64_t> A(N);
  for (int i = 0; i < N; ++i) cin >> A[i];

  int64_t P = A[0] + A[1];
  int64_t S = 0; for (int i = 2; i < N; ++i) S ^= A[i];

  if (P - S & 1) {
    cout << -1 << endl;
    return 0;
  }

  int64_t D = P - S >> 1;

  int64_t a = D;
  int64_t b = D;
  for (int i = 40; ~i; --i) if (S >> i & 1) {
    int64_t na = a | (int64_t) 1 << i;
    if (na <= A[0]) a = na;
    else b = b | (int64_t) 1 << i;
  }

  if (a + b == P && (a ^ b) == S && (1 <= a && a <= A[0])) {
    cout << A[0] - a << endl;
  } else {
    cout << -1 << endl;
  }

  return 0;
}


```