/*====Corycle====*/
#include<algorithm>
#include<iostream>
#include<iomanip>
#include<cstring>
#include<cstdlib>
#include<complex>
#include<cstdio>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<map>
#define ll long long
using namespace std;
const int inf=0x3f3f3f3f;
const ll N=305;
ll read(){
	ll s=0,f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){s=s*10+c-'0';c=getchar();}
	return s*f;
}
ll n,a[N],sum,tot,ans;
int main(){
	n=read();
	for(ll i=1;i<=n;i++)a[i]=read();
	for(ll i=3;i<=n;i++)sum^=a[i];
	tot=a[1]+a[2];tot-=sum;
	if(tot<0){puts("-1");return 0;}
	for(ll i=42;i>=0;i--){
		if(2*(1ll<<i)<=tot){
			if((sum>>i&1)){puts("-1");return 0;}
			ans+=(1ll<<i);
			tot-=2*(1ll<<i);
		}
	}
	if(tot){puts("-1");return 0;}
	if(ans>a[1]){puts("-1");return 0;}
	for(ll i=42;i>=0;i--){
		if((sum>>i&1)&&(ans+(1ll<<i))<=a[1])ans+=(1ll<<i);
	}
	if(ans==0){puts("-1");return 0;}
	cout<<a[1]-ans;
	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
/*====Corycle====*/
#include<algorithm>
#include<iostream>
#include<iomanip>
#include<cstring>
#include<cstdlib>
#include<complex>
#include<cstdio>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<map>
#define ll long long
using namespace std;
const int inf=0x3f3f3f3f;
const ll N=305;
ll read(){
	ll s=0,f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){s=s*10+c-'0';c=getchar();}
	return s*f;
}
ll n,a[N],sum,tot,ans;
int main(){
	n=read();
	for(ll i=1;i<=n;i++)a[i]=read();
	for(ll i=3;i<=n;i++)sum^=a[i];
	tot=a[1]+a[2];tot-=sum;
	if(tot<0){puts("-1");return 0;}
	for(ll i=42;i>=0;i--){
		if(2*(1ll<<i)<=tot){
			if((sum>>i&1)){puts("-1");return 0;}
			ans+=(1ll<<i);
			tot-=2*(1ll<<i);
		}
	}
	if(tot){puts("-1");return 0;}
	if(ans>a[1]){puts("-1");return 0;}
	for(ll i=42;i>=0;i--){
		if((sum>>i&1)&&(ans+(1ll<<i))<=a[1])ans+=(1ll<<i);
	}
	if(ans==0){puts("-1");return 0;}
	cout<<a[1]-ans;
	return 0;
}
```