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

void fail(){
    cout << -1 << endl;
    exit(0);
}

int nth_bit(int64_t num, int n){
    return (num >> n) & 1;
}

int main(){
    int N;
    cin >> N;
    vector<int64_t> A(3);
    for(int i=0; i<N; i++){
        int64_t a;
        cin >> a;
        A[min(i, 2)] ^= a;
    }
    int64_t ans = 0;
    for(int k=40; k>=0; k--){
        int64_t d = 1LL<<k;
        if((nth_bit(A[0], k)^nth_bit(A[1], k)) != nth_bit(A[2], k)){
            int64_t v = min(A[0]%d+1, d - A[1]%d);
            if(A[0] < v) fail();
            A[0] -= v;
            A[1] += v;
            k = 41;
            ans += v;
        }
    }
    if(A[0] == 0) fail();
    cout << ans << endl;
    return 0;
}

### Prompt

Create a solution in cpp for 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;

void fail(){
    cout << -1 << endl;
    exit(0);
}

int nth_bit(int64_t num, int n){
    return (num >> n) & 1;
}

int main(){
    int N;
    cin >> N;
    vector<int64_t> A(3);
    for(int i=0; i<N; i++){
        int64_t a;
        cin >> a;
        A[min(i, 2)] ^= a;
    }
    int64_t ans = 0;
    for(int k=40; k>=0; k--){
        int64_t d = 1LL<<k;
        if((nth_bit(A[0], k)^nth_bit(A[1], k)) != nth_bit(A[2], k)){
            int64_t v = min(A[0]%d+1, d - A[1]%d);
            if(A[0] < v) fail();
            A[0] -= v;
            A[1] += v;
            k = 41;
            ans += v;
        }
    }
    if(A[0] == 0) fail();
    cout << ans << endl;
    return 0;
}

```