John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces nn chips today, the ii-th chip produced this day has a serial number sisi. 

At the end of the day, he packages all the chips produced this day, and send it to wholesalers. More specially, he writes a checksum number on the package, this checksum is defined as below: 

maxi,j,k(si+sj)⊕skmaxi,j,k(si+sj)⊕sk

which i,j,ki,j,k are three different integers between 11 and nn. And ⊕⊕ is symbol of bitwise XOR. 

Can you help John calculate the checksum number of today?

Input

The first line of input contains an integer TT indicating the total number of test cases. 

The first line of each test case is an integer nn, indicating the number of chips produced today. The next line has nn integers s1,s2,..,sns1,s2,..,sn, separated with single space, indicating serial number of each chip. 

1≤T≤10001≤T≤1000 
3≤n≤10003≤n≤1000 
0≤si≤1090≤si≤109 
There are at most 1010 testcases with n>100n>100

Output

For each test case, please output an integer indicating the checksum number in a line.

Sample Input

2
3
1 2 3
3
100 200 300

Sample Output 

6
400

 字典樹求異或值的時候，先把每個數按照二進位制從高位到地位存到樹裡面，

查詢的時候，從最高位開始找相反的結點（0找1，1找0）

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define inf 0x3f3f3f
using namespace std;
#define mod 1000000007
const int maxn=100005;
ll a[maxn];
struct node{
    node *son[2];
    ll cut,val;
    node(){
    cut=0;
    val=0;
    son[0]=son[1]=NULL;
    }
};
node *root;
void add(ll x){
    node *p=root,*q;
    for(int i=31;i>=0;i--){
     int y=((x&(1<<i))!=0);
     if(p->son[y]!=NULL){
        p=p->son[y];
     }
     else{
        q=new node;
        p->son[y]=q;
            p=q;
        }
        p->val++;
    }
    p->cut=x;
}
void del(ll x){
    node *p=root;
    for(int i=31;i>=0;i--){
     int y=((x&(1<<i))!=0);
        p=p->son[y];
        p->val--;
    }
}
ll fin(ll x,ll a,ll b){
    node *p=root;
    for(int i=31;i>=0;i--){
        int y=((x&(1<<i))!=0);
        if(p->son[!y]&&p->son[!y]->val){
            p=p->son[!y];}
        else
            p=p->son[y];
    }
    return x^(p->cut);
}
int main(){
    int t;
    scanf("%d",&t);
    for(int I=1;I<=t;I++){
        int n,m;
        root=new node;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%lld",&a[i]);
            add(a[i]);
        }
        ll ans=-1;
        for(int i=0;i<n;i++){
            del(a[i]);
            for(int j=i+1;j<n;j++){
                if(i!=j){
                    del(a[j]);
                    ans=max(ans,fin(a[i]+a[j],a[i],a[j]));
                    add(a[j]);
                }
            }
            add(a[i]);
       }
        printf("%lld\n",ans);
    }
}