Educational Codeforces Round 83 (Rated for Div. 2) B. Bogosort（排序/思维）

You are given an array a1,a2,…,ana1,a2,…,an . Array is good if for each pair of indexes i<ji<j the condition j−aj≠i−aij−aj≠i−ai holds. Can you shuffle this array so that it becomes good? To shuffle an array means to reorder its elements arbitrarily (leaving the initial order is also an option).

For example, if a=[1,1,3,5]a=[1,1,3,5] , then shuffled arrays [1,3,5,1][1,3,5,1] , [3,5,1,1][3,5,1,1] and [5,3,1,1][5,3,1,1] are good, but shuffled arrays [3,1,5,1][3,1,5,1] , [1,1,3,5][1,1,3,5] and [1,1,5,3][1,1,5,3] aren't.

It's guaranteed that it's always possible to shuffle an array to meet this condition.

Input

The first line contains one integer tt (1≤t≤1001≤t≤100 ) — the number of test cases.

The first line of each test case contains one integer nn (1≤n≤1001≤n≤100 ) — the length of array aa .

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100 ).

Output

For each test case print the shuffled version of the array aa which is good.

Example

Input

Copy

3
1
7
4
1 1 3 5
6
3 2 1 5 6 4

Output

Copy

7
1 5 1 3
2 4 6 1 3 5
调换一下不等号两边的项，变成了j-i≠aj-ai,这样就好办了，因为注意到输出任意序列即可，而i要小于j，只需要把原序列降序排列，就能使j-i>0而aj-ai<=0。

#include <bits/stdc++.h>
using namespace std;
int n,i,a[105];
bool cmp(int a, int b)
{
    return a>b;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a+1,a+n+1,cmp);
        for(i=1;i<=n;i++)
        {
            printf("%d ",a[i]);
        }
        cout<<endl;
    }
    return 0;
}