E - Stoned Game

T is playing a game with his friend, HL.

There are nn piles of stones, the ii-th pile initially has a_iai​ stones.

T and HL will take alternating turns, with T going first. In each turn, a player chooses a non-empty pile and then removes a single stone from it. However, one cannot choose a pile that has been chosen in the previous turn (the pile that was chosen by the other player, or if the current turn is the first turn then the player can choose any non-empty pile). The player who cannot choose a pile in his turn loses, and the game ends.

Assuming both players play optimally, given the starting configuration of tt games, determine the winner of each game.

Input

The first line of the input contains a single integer tt (1 \le t \le 100)(1≤t≤100) — the number of games. The description of the games follows. Each description contains two lines:

The first line contains a single integer nn (1 \le n \le 100)(1≤n≤100) — the number of piles.

The second line contains nn integers a_1, a_2, \dots, a_na1​,a2​,…,an​ (1 \le a_i \le 100)(1≤ai​≤100).

Output

For each game, print on a single line the name of the winner, "T" or "HL" (without quotes)

Sample 1

2
1
2
2
1 1

T
HL

Note

In the first game, T removes a single stone from the only pile in his first turn. After that, although the pile still contains 11 stone, HL cannot choose from this pile because it has been chosen by T in the previous turn. Therefore, T is the winner.

#include<iostream>
#include<algorithm>
#include<map>
#include<cstring>
#include<vector>
#include<cmath>
#include<cstdlib>
 
using namespace std;
const int N = 1e5 + 10;
 
int a[N], b[N];
 
//简化成两种情况即可：
//1.存在一堆石头比剩下的几堆加起来都要多，这样T先手就可以一直取最多的那堆，直到赢
//2.全部石头加一起为奇数，T赢，偶数，T输
 
int main()
{
	int t; cin >> t;
	while (t--)
	{
		int n; cin >> n;
		for (int i = 1; i <= n; i++) cin >> a[i];
		
		sort(a + 1, a + n + 1);
 
		int sum = 0;
 
		for (int i = 1; i < n; i++) sum += a[i];
 
		if (a[n] > sum)
		{
			cout << "T" << endl;
			continue;
		}
		else
		{
			sum += a[n];
			if (sum & 1) cout << "T" << endl;
			else cout << "HL" << endl;
		}
	}
}