#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=(1LL<<45);
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;


int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n, x, a, b, sum = 0;
    cin >> n;
    for(int i=0;i<n;i++){
        if(i == 0) cin >> a; 
        else if(i == 1) cin >> b;
        else {
        	cin >> x;
        	sum ^= x;
        }
    }
    ll num = a + b - sum;
    if(num < 0LL || (num & 1)){
    	cout << -1 << endl;
    	return 0;
    }
    num /= 2LL;
    vector<ll> c(45, -1LL);
    for(int i=0;i<45;i++) if((sum >> i & 1) == 0){
    	if(num >> i & 1) c[i] = 1;
    	else c[i] = 0;
    }
    ll ans = 0;
    for(int i=0;i<45;i++) if(c[i] == 1LL) ans += 1LL << i;
    for(int i=44;i>=0;i--) if(c[i] == -1LL){
        if(ans + (1LL << i) <= a) ans += 1LL << i;
    }
    num = b + a - ans;

    if(ans == 0LL || ans > a || (num ^ ans) != sum) cout << -1 << endl;
    else cout << a - ans << 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>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=(1LL<<45);
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;


int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n, x, a, b, sum = 0;
    cin >> n;
    for(int i=0;i<n;i++){
        if(i == 0) cin >> a; 
        else if(i == 1) cin >> b;
        else {
        	cin >> x;
        	sum ^= x;
        }
    }
    ll num = a + b - sum;
    if(num < 0LL || (num & 1)){
    	cout << -1 << endl;
    	return 0;
    }
    num /= 2LL;
    vector<ll> c(45, -1LL);
    for(int i=0;i<45;i++) if((sum >> i & 1) == 0){
    	if(num >> i & 1) c[i] = 1;
    	else c[i] = 0;
    }
    ll ans = 0;
    for(int i=0;i<45;i++) if(c[i] == 1LL) ans += 1LL << i;
    for(int i=44;i>=0;i--) if(c[i] == -1LL){
        if(ans + (1LL << i) <= a) ans += 1LL << i;
    }
    num = b + a - ans;

    if(ans == 0LL || ans > a || (num ^ ans) != sum) cout << -1 << endl;
    else cout << a - ans << endl;
    return 0;
}

```