/*
    1 2 3 4 5 6 7 8 9 10 11 
    5 6

    4 6
    3 6
    3 5
    3 4
    2 4
    1 4
    1 3
    1 3
    0 3

    4 6
    4 5
    4 4
    3 4
    2 4
    2 3
    2 2
    1 2
    0 2

    2 2 3 2 2 

    od = 1
    ev = 4

    1 4


*/

/* Solved by Sakhawat Hossain Mahin, CSE-28th, SMUCT */
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long

#define faf ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define test  \
    int T;    \
    cin >> T; \
    while (T--)

const int N = 1e5 + 7;
int ar[N];

void crack()
{
    int n;
    cin>>n;
    int od = 0, ev = 0; 
    for(int i = 0; i < n; i++){
        int x;
        cin>>x;
        if(x % 2){
            od++;
        }
        else ev++;
    }
    if(od == ev){
        cout << "Hridoy" << endl;
        return;
    }
    int g = (od <= ev ? od : ev);

    if(g % 2){
        cout << "Roy" << endl;
    }
    else cout << "Hridoy" << endl;

}

int32_t main()
{
    faf auto st = clock();
    test
    crack();
    cerr << 1.0 * (clock() - st) / CLOCKS_PER_SEC << endl;
    return 0;
}

/*
    od = 5, ev = 5;
    roy hridoy
r    4 5
h    4 4
r    4 3
h    3 3
r    2 3
h    2 2
r    2 1
h    1 1
r    0 1
h    0 0
r
h wins
*/