#include <bits/stdc++.h>

using namespace std;
using ll=long long;

void solve() {
    int n; cin >> n;
    ll A,B,C = 0;
    cin >> A >> B;
    for (int i = 3; i <= n; i++) {
        ll x; cin >> x;
        C ^= x;
    }
    ll S = A + B;
    if (S<C || ((S-C)&1)) {
        cout << -1; return;
    }
    ll D = (S-C)/2;
    if (D&C) {
        cout << -1; return;
    }
    if (D > A) {
        cout << -1; return;
    }
    ll a = 0;
    for (int i = 60; i >= 0; i--) {
        if (C>>i&1) {
            if (a+(1ll<<i)+D <= A) a += 1ll<<i;
        }
    }
    ll res = A-D-a;
    if (res == A) {
        cout << -1; return;
    }
    cout << res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    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;
using ll=long long;

void solve() {
    int n; cin >> n;
    ll A,B,C = 0;
    cin >> A >> B;
    for (int i = 3; i <= n; i++) {
        ll x; cin >> x;
        C ^= x;
    }
    ll S = A + B;
    if (S<C || ((S-C)&1)) {
        cout << -1; return;
    }
    ll D = (S-C)/2;
    if (D&C) {
        cout << -1; return;
    }
    if (D > A) {
        cout << -1; return;
    }
    ll a = 0;
    for (int i = 60; i >= 0; i--) {
        if (C>>i&1) {
            if (a+(1ll<<i)+D <= A) a += 1ll<<i;
        }
    }
    ll res = A-D-a;
    if (res == A) {
        cout << -1; return;
    }
    cout << res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}

```