A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.

For example, the following four sequences are growing: 

  * [2, 3, 15, 175] — in binary it's [10_2, 11_2, 1111_2, 10101111_2]; 
  * [5] — in binary it's [101_2]; 
  * [1, 3, 7, 15] — in binary it's [1_2, 11_2, 111_2, 1111_2]; 
  * [0, 0, 0] — in binary it's [0_2, 0_2, 0_2]. 



The following three sequences are non-growing: 

  * [3, 4, 5] — in binary it's [11_2, 100_2, 101_2]; 
  * [5, 4, 3] — in binary it's [101_2, 100_2, 011_2]; 
  * [1, 2, 4, 8] — in binary it's [0001_2, 0010_2, 0100_2, 1000_2]. 



Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 ⊕ y_1, x_2 ⊕ y_2, ..., x_n ⊕ y_n is growing where ⊕ denotes [bitwise XOR](http://tiny.cc/bry9uz).

You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.

The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 ≤ k ≤ n such that a_i = b_i for any 1 ≤ i < k but a_k < b_k.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — length of the sequence x_i.

The second line contains n integers x_1, x_2, ..., x_n (0 ≤ x_i < 2^{30}) — elements of the sequence x_i.

It is guaranteed that the sum of n overall all test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print n integers y_1, y_2, ..., y_n (0 ≤ y_i < 2^{30}) — lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.

Example

Input


5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0


Output


0 0 0 0 
0 1 3 7 
0 1 0 3 2 
0 2 0 14 
0 
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define ld long double
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.14159265
ll const mod1=1000000007;
// bool comp(pair<ll,ll>a,pair<ll,ll>b)
// {
//    if(a.first==b.first)
//     return a.second<b.second;
//   // else
//     return a.first>b.first;

// }
// sort(a.begin(),a.end(),greater<int>());
// priority_queue<int, vector<int>, greater<int> > Q;
// vector<ll> factorial(1001,1);
// vector<ll> inv_factorial(1001);
// vector<vector<ll> > combi(1001,vector<ll>(1001));
// ll binpow(ll a, ll b)
// {
//     ll res = 1;
//     while (b > 0) {
//         if (b & 1)
//             res = res * a;
//         a = a * a;
//         b >>= 1;
//     }
//     return res;
// }
 
// ll binpow(ll a, ll b, ll mod)
// {
//     ll res = 1;
//     while (b > 0) {
//         if (b & 1)
//             res = (res * a)%mod;
//         a = (a * a)%mod;
//         b >>= 1;
//     }
//     return res%mod;
// }
 
// ll gcd(ll a, ll b)
// {
//     if(b==0) return a;
//     else return gcd(b,a%b);
// }
 
// ll lcm (ll a, ll b)
// {
//     return ((a / gcd(a, b)) * b);
// }
 
// bool comp(ll a, ll b)
// {
//     return a>b;
// }
 
// ll inversemod(ll a, ll mod)
// {
//     return binpow(a,mod-2, mod);
// }
// for(int i=1;i<=1000;i++) factorial[i]=(factorial[i-1]*i)%mod1;
 
//   for(int i=0;i<=1000;i++) inv_factorial[i]=inversemod(factorial[i],mod1);
 
//   for(int i=0;i<=1000;i++)
//   {
//     for(int j=0;j<=i;j++)
//     {
//       combi[i][j]=(((factorial[i]*inv_factorial[i-j])%mod1)*inv_factorial[j])%mod1;
//     }
//   }
// ll t[4*20001]
// void build(ll a[], ll v, ll tl, ll tr) {
//     if (tl == tr) {
//         t[v] = a[tl];
//     } else {
//         int tm = (tl + tr) / 2;
//         build(a, v*2, tl, tm);
//         build(a, v*2+1, tm+1, tr);
//         t[v] = t[v*2] ^ t[v*2+1];
//     }
// }
// void dfs(ll i,ll *visited,vector<ll> e[])
// {
//   visited[i]=1;
//   ll n=e[i].size();
//   for(ll j=0;j<n;j++)
//   {
//     if(visited[e[i][j]]==0)
//       dfs(e[i][j],visited,e);
//   }
// }
// ld ans=0;

int main()
{
   ios_base::sync_with_stdio(false);
   cin.tie(0);
   ll test;
   cin>>test;
//    test=1;
   for(ll ftest=0;ftest<test;ftest++)
   {
      int n,i;
      cin>>n;
      vector<ll>A(n),ans(n);
      for(i=0;i<n;i++)cin>>A[i];
      ans[0]=0;
      ll prev=A[0];
      for(i=1;i<n;i++)
      {
          ll b=A[i]&prev;
          ll y=b^prev;
          ans[i]=y;
          prev=A[i]^y;
      }
      for(i=0;i<n;i++)cout<<ans[i]<<' ';
      cout<<'\n';
   }
}