#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[305];
int n;
ll x;
ll sum;
int main(){
    scanf("%d",&n);
    for (int i = 0; i < n; i++){
        scanf("%lld",&a[i]);
        if (i >= 2) x ^= a[i];
    }
    sum = a[0] + a[1];
    if (sum < x || ((sum-x)&1ll)){
        printf("-1\n");
        return 0;
    }
    ll N = (sum-x)>>1ll;
    //printf("sum = %lld, xor = %lld, thus and = %lld\n",sum,x,N);
    if (N > a[0]){
        printf("-1\n");
        return 0;
    }
    ll cur = N;
    for (int i = 60; i >= 0; i--){
        ll NEW = cur|(1ll<<i);
        if (NEW <= a[0] && (NEW^(sum-NEW)) == x ){
            //printf("can on %d\n",i);
            cur = NEW;
            //printf("check: %lld vs %lld\n",cur^(sum-cur),x);
        }
    }
    if (cur == 0){
        printf("-1\n");
        return 0;
    }
    ll ans = a[0]-cur;
    if (cur <= a[0] && (cur^(sum-cur)) == x){
        printf("%lld\n",ans);
    }
    else printf("-1\n");
}

### Prompt

Your task is to create a CPP solution to 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;
typedef long long ll;
ll a[305];
int n;
ll x;
ll sum;
int main(){
    scanf("%d",&n);
    for (int i = 0; i < n; i++){
        scanf("%lld",&a[i]);
        if (i >= 2) x ^= a[i];
    }
    sum = a[0] + a[1];
    if (sum < x || ((sum-x)&1ll)){
        printf("-1\n");
        return 0;
    }
    ll N = (sum-x)>>1ll;
    //printf("sum = %lld, xor = %lld, thus and = %lld\n",sum,x,N);
    if (N > a[0]){
        printf("-1\n");
        return 0;
    }
    ll cur = N;
    for (int i = 60; i >= 0; i--){
        ll NEW = cur|(1ll<<i);
        if (NEW <= a[0] && (NEW^(sum-NEW)) == x ){
            //printf("can on %d\n",i);
            cur = NEW;
            //printf("check: %lld vs %lld\n",cur^(sum-cur),x);
        }
    }
    if (cur == 0){
        printf("-1\n");
        return 0;
    }
    ll ans = a[0]-cur;
    if (cur <= a[0] && (cur^(sum-cur)) == x){
        printf("%lld\n",ans);
    }
    else printf("-1\n");
}

```