#include<bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin >> N;
  vector<int64_t> vec(N);
  for(int i=0; i<N; i++){
    cin >> vec[i];
  }
  int64_t sum = 0;
  for(int i=2; i<N; i++){
    sum ^= vec[i];
  }
  
  int64_t two = vec[0] + vec[1];
  
  bool flag = true;
  if(two < sum){
    flag = false;
  }
  
  int64_t A = 0;
  int64_t a = vec[0];
  int64_t B = 0;
  two -= sum;
  
  bitset<60> S(sum);
  bitset<60> T(two);
  
  for(int64_t i=0; i<59; i++){
    if(T.test(i)){
      if(i == 0){
        flag = false;
      }
      else if(S.test(i-1)){
        flag = false;
      }
      else{
        int64_t X = pow(2,i-1);
        A += X;
      }
    }
  }
  
  a = vec[0] - A;
  
  for(int64_t i=59; i>=0; i--){
    if(S.test(i)){
      int64_t X = pow(2,i);
      if(a >= X){
        A += X;
        a -= X;
      }
    }
  }
  
  
  if(!flag || A > vec[0] || A == 0){
    cout << -1 << endl;
  }
  else{
    cout << vec[0] - A << endl;
  }
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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(){
  int N;
  cin >> N;
  vector<int64_t> vec(N);
  for(int i=0; i<N; i++){
    cin >> vec[i];
  }
  int64_t sum = 0;
  for(int i=2; i<N; i++){
    sum ^= vec[i];
  }
  
  int64_t two = vec[0] + vec[1];
  
  bool flag = true;
  if(two < sum){
    flag = false;
  }
  
  int64_t A = 0;
  int64_t a = vec[0];
  int64_t B = 0;
  two -= sum;
  
  bitset<60> S(sum);
  bitset<60> T(two);
  
  for(int64_t i=0; i<59; i++){
    if(T.test(i)){
      if(i == 0){
        flag = false;
      }
      else if(S.test(i-1)){
        flag = false;
      }
      else{
        int64_t X = pow(2,i-1);
        A += X;
      }
    }
  }
  
  a = vec[0] - A;
  
  for(int64_t i=59; i>=0; i--){
    if(S.test(i)){
      int64_t X = pow(2,i);
      if(a >= X){
        A += X;
        a -= X;
      }
    }
  }
  
  
  if(!flag || A > vec[0] || A == 0){
    cout << -1 << endl;
  }
  else{
    cout << vec[0] - A << endl;
  }
}
```