You are given an integer n and an array a_1, a_2, …, a_n. You should reorder the elements of the array a in such way that the sum of MEX on prefixes (i-th prefix is a_1, a_2, …, a_i) is maximized.

Formally, you should find an array b_1, b_2, …, b_n, such that the sets of elements of arrays a and b are equal (it is equivalent to array b can be found as an array a with some reordering of its elements) and ∑_{i=1}^{n} MEX(b_1, b_2, …, b_i) is maximized.

MEX of a set of nonnegative integers is the minimal nonnegative integer such that it is not in the set.

For example, MEX(\{1, 2, 3\}) = 0, MEX(\{0, 1, 2, 4, 5\}) = 3.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 100).

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100).

Output

For each test case print an array b_1, b_2, …, b_n — the optimal reordering of a_1, a_2, …, a_n, so the sum of MEX on its prefixes is maximized.

If there exist multiple optimal answers you can find any.

Example

Input


3
7
4 2 0 1 3 3 7
5
2 2 8 6 9
1
0


Output


0 1 2 3 4 7 3 
2 6 8 9 2 
0 

Note

In the first test case in the answer MEX for prefixes will be: 

  1. MEX(\{0\}) = 1 
  2. MEX(\{0, 1\}) = 2 
  3. MEX(\{0, 1, 2\}) = 3 
  4. MEX(\{0, 1, 2, 3\}) = 4 
  5. MEX(\{0, 1, 2, 3, 4\}) = 5 
  6. MEX(\{0, 1, 2, 3, 4, 7\}) = 5 
  7. MEX(\{0, 1, 2, 3, 4, 7, 3\}) = 5 

The sum of MEX = 1 + 2 + 3 + 4 + 5 + 5 + 5 = 25. It can be proven, that it is a maximum possible sum of MEX on prefixes.
#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define int long long
#define double long double
#define pb push_back
#define mp make_pair
#define endl '\n'
#define all(x) x.begin(),x.end()
#define fill(a,b) memset(a,b,sizeof(a))
#define sz(x) (int)x.size()
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define deb(x) cout<<#x<<" "<<x<<endl;
#define T int t; cin>>t; while(t--)
#define fst ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

int firstMissingPositive(int arr[], int n)
{
     
    for (int i = 0; i < n; i++) {

        while (arr[i] >= 1 && arr[i] <= n
               && arr[i] != arr[arr[i] - 1]) {
            swap(arr[i], arr[arr[i] - 1]);
        }
    }

    for (int i = 0; i < n; i++) {
        if (arr[i] != i + 1) {
            return i + 1;
        }
    }
   
    return n + 1;
}

signed main()
{

    fst;

#ifndef ONLINE_JUDGE
    freopen("input3.txt", "r", stdin);
    freopen("output3.txt", "w", stdout);
#endif

    
    T{
        int n,p,q, m,h, k, sum = 0, cnt=0, x, y, z, flag = 0, pos, diff,r=0,l=0,u=0,d=0,cnt1=0,cnt0=0,cnt2=0,ans,mi=INT_MAX,mx=INT_MIN;
        cin>>n;
        
        int a[n];
        vector<int> v,v1,v2;
        set<int> st;
        map<int,int> mp;

        rep(i,0,n) {
            cin>>x;
            if(find(all(st),x) != st.end())
              v.pb(x);
            st.insert(x);
        }

        for(auto i:st) cout<<i<<" ";
        sort(all(v));

        for(auto i:v) cout<<i<<" ";


        cout<<endl;

        v.clear();
        st.clear();

    }
}
