Diablo III is an action role-playing video game. A few days ago, Reaper of Souls (ROS), the new expansion of Diablo III, has been released! On hearing the news, the crazy video game nerd Yuzhi shouted: "I'm so excited! I'm so excited! I wanna kill the Diablo once more!"

The ROS introduced a lot of new features and changes. For example, there are two new attributes for players in the game: Damage and Toughness. The attribute Damage indicates the amount of damage per second you can deal and the Toughness is the total amount of raw damage you can take.

To beat the Diablo, Yuzhi need to select the most suitable equipments for himself. A player can carry at most 13 equipments in 13 slots: Head, Shoulder, Neck, Torso, Hand, Wrist, Waist, Legs, Feet, Shield, Weapon and 2 Fingers. By the way, there is a special type of equipment: Two-Handed. A Two-Handed equipment will occupy both Weapon and Shield slots.

Each equipment has different properties on Damage and Toughness, such as a glove labeled "30 20" means that it can increase 30 Damage and 20 Toughness for the player who equips it in the Hand slot. The total Damage and Toughness is the sum of Damage and Toughness of all equipments on the body. A player without any equipments has 0 Damage and 0 Toughness.

Yuzhi has N equipments stored in his stash. To fight against the Diablo without lose the battle, he must have at least M Toughness. In addition, he want to finish the battle as soon as possible. That means the Damage should be as much as possible. Please help Yuzhi to determine which equipments he should take.

Input

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:

The first line contains 2 integers N (1 <= N <= 300) and M (0 <= M <= 50000). The next N lines are the description of equipments. The i-th line contains a string Siand two integers Di and Ti (1 <= Di, Ti <= 50000). Si is the type of equipment in {"Head", "Shoulder", "Neck", "Torso", "Hand", "Wrist", "Waist", "Legs", "Feet", "Finger", "Shield", "Weapon", "Two-Handed"}. Di and Ti are the Damage and Toughness of this equipment.

Output

For each test case, output the maximum Damage that Yuzhi can get, or -1 if he can not reach the required Toughness.

Sample Input

2
1 25
Hand 30 20
5 25
Weapon 15 5
Shield 5 15
Two-Handed 25 5
Finger 5 10
Finger 5 10

Sample Output

-1
35

       每個位置上有一堆物品，求最佳的組合方式使防禦力大於等於m的情況下攻擊力最大。

       如果沒有限制條件的話這個就是一個普通的dp，但是存在著兩個特殊請況：：

       1.如果選擇（雙手武器），那麼就不能拿雙手上的武器（劍和盾？）。

       2.還有就是戒指可以帶兩個。

       這樣的話我們可以將這兩種情況下的所有可能分別化為獨特的兩組集進行dp就好了（所有可能的武器選擇作為一組，所有可能的戒指裝戴情況算作一組）

#include <bits/stdc++.h>
using namespace std;
const int maxn = 15;
const int maxm = 60000;
struct fuck {
	int atk, def;
	fuck(int a, int b) {
		this->atk = a; this->def = b;
	}
};
map<string, int>mp;
int sum[maxn];
vector<fuck>G[maxn];
int n, m, dp[maxn][maxm];
bool cmp(int a, int b) {
	return G[a].size() > G[b].size();
}
void init() {
	mp["Weapon"] = 1;mp["Shield"] = 2;mp["Two-Handed"] = 3;
	mp["Finger"] = 4;mp["Feet"] = 5;mp["Legs"] = 6;mp["Waist"] = 7;
	mp["Wrist"] = 8;mp["Hand"] = 9;mp["Torso"] = 10;
	mp["Neck"] = 11;mp["Shoulder"] = 12;mp["Head"] = 13;
	for (int s = 1; s <= 13; s++)
		sum[s] = s;
}
void solve() {
	dp[2][0] = 0;
	for (int s = 3; s <= 13; s++) {
		int x = sum[s];
		for (int i = m; i >= 0; i--) {
			dp[s][i] = max(dp[s][i], dp[s - 1][i]);
			if (dp[s - 1][i] == -1)continue;
			for (int j = 0; j < G[x].size(); j++) {
				fuck t = G[x][j];
				int d = min(m, i + t.def);
				dp[s][d] = max(dp[s][d], dp[s - 1][i] + t.atk);
			}
		}
	}
}
int main() {
	init();
	int te; scanf("%d", &te);
	while (te--) {
		memset(dp, -1, sizeof dp);
		for (int i = 1; i <= 13; i++) G[i].clear();
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) {
			char m[20]; scanf("%s", m);
			int v = mp[m];
			int a, b; scanf("%d%d", &a, &b);
			G[v].push_back(fuck(a, b));
		}
		for (int i = 1; i <= 2; i++)
			for (int j = 0; j < G[i].size(); j++)
				G[3].push_back(G[i][j]);
		for (int i = 0; i < G[1].size(); i++)
			for (int j = 0; j < G[2].size(); j++)
				G[3].push_back(fuck(G[1][i].atk + G[2][j].atk, G[1][i].def + G[2][j].def));
		int t = G[4].size();
		for (int i = 0; i < t; i++) {
			for (int j = i + 1; j < t; j++) {
				if (i == j)continue;
				G[4].push_back(fuck(G[4][i].atk + G[4][j].atk, G[4][i].def + G[4][j].def));
			}
		}
		sort(sum + 3, sum + 14, cmp);
		solve();
		cout << dp[13][m] << endl;
	}
}
/*
	2
	5 25
	Weapon 15 5
	Shield 5 15
	Two-Handed 25 5
	Finger 5 10
	Finger 5 10
*/