// Arup Guha
// 4/13/2020
// Alternate Solution to COP 4516 Team Final Contest Problem: Elf Employment

import java.util.*;

public class elf_arup {

	public static int nToys;
	public static int nKids;
	public static boolean[][] likes;
	public static boolean[] good;

	public static void main(String[] args) {
	
		Scanner stdin = new Scanner(System.in);
		int nC = stdin.nextInt();
		
		// Process cases.
		for (int loop=0; loop<nC; loop++) {
		
			nToys = stdin.nextInt();
			nKids = stdin.nextInt();
			likes = new boolean[nKids][nToys];
			good = new boolean[nKids];
			
			// Read in each kid.
			for (int i=0; i<nKids; i++) {
			
				// See if the are good or not!
				good[i] = stdin.next().equals("nice");
			
				// Set to true all the toys this kid likes.
				int nLike = stdin.nextInt();
				for (int j=0; j<nLike; j++)
					likes[i][stdin.nextInt()-1] = true;
			}
			
			// Safe bounds for the binary search.
			int low = 1, high = nKids;
			
			// Run binary search.
			while (low < high) {
			
				int mid = (low+high)/2;
				
				// We can do it, so answer is no more than mid.
				if (canDo(mid))
					high = mid;
					
				// We can't so it has to be at least mid+1.
				else
					low = mid+1;
			}
			
			// Ta da!
			System.out.println(low);
		}
	}
	
	public static boolean canDo(int nDays) {
	
		int n = nToys + nKids;
		Dinic graph = new Dinic(n);
		
		// We make up to nDays of each toy.
		for (int i=0; i<nToys; i++)
			graph.add(n,i,nDays,0);
			
		// Each kid just needs one toy.
		for (int i=0; i<nKids; i++)
			graph.add(nToys+i,n+1,1,0);
		
		// Only allow each kid to get the type of toy they are allowed in the problem.
		for (int i=0; i<nKids; i++) {
			for (int j=0; j<nToys; j++) {
			
				// Kid i can't get toy j...
				if (good[i] != likes[i][j]) continue;
			
				// Kid i is allowed to get toy i.
				graph.add(j, nToys+i, 1, 0);
			}
		}
		
		// We're good if each kid gets 1 toy.
		return graph.flow() == nKids;
	}
}

/*** UCF Dinic Code ***/
class Edge {
	int v1, v2, cap, flow;
	Edge rev;
	Edge(int V1, int V2, int Cap, int Flow) {
		v1 = V1;
		v2 = V2;
		cap = Cap;
		flow = Flow;
	}
}

class Dinic {

	// Queue for the top level BFS.
	public ArrayDeque<Integer> q;

	// Stores the graph.
	public ArrayList<Edge>[] adj;
	public int n;

	// s = source, t = sink
	public int s;
	public int t;


	// For BFS.
	public boolean[] blocked;
	public int[] dist;

	final public static int oo = (int)1E9;

	// Constructor.
	public Dinic (int N) {

		// s is the source, t is the sink, add these as last two nodes.
		n = N; s = n++; t = n++;

		// Everything else is empty.
		blocked = new boolean[n];
		dist = new int[n];
		q = new ArrayDeque<Integer>();
		adj = new ArrayList[n];
		for(int i = 0; i < n; ++i)
			adj[i] = new ArrayList<Edge>();
	}

	// Just adds an edge and ALSO adds it going backwards.
	public void add(int v1, int v2, int cap, int flow) {
		Edge e = new Edge(v1, v2, cap, flow);
		Edge rev = new Edge(v2, v1, 0, 0);
		adj[v1].add(rev.rev = e);
		adj[v2].add(e.rev = rev);
	}

	// Runs other level BFS.
	public boolean bfs() {

		// Set up BFS
		q.clear();
		Arrays.fill(dist, -1);
		dist[t] = 0;
		q.add(t);

		// Go backwards from sink looking for source.
		// We just care to mark distances left to the sink.
		while(!q.isEmpty()) {
			int node = q.poll();
			if(node == s)
				return true;
			for(Edge e : adj[node]) {
				if(e.rev.cap > e.rev.flow && dist[e.v2] == -1) {
					dist[e.v2] = dist[node] + 1;
					q.add(e.v2);
				}
			}
		}

		// Augmenting paths exist iff we made it back to the source.
		return dist[s] != -1;
	}

	// Runs inner DFS in Dinic's, from node pos with a flow of min.
	public int dfs(int pos, int min) {

		// Made it to the sink, we're good, return this as our max flow for the augmenting path.
		if(pos == t)
			return min;
		int flow = 0;

		// Try each edge from here.
		for(Edge e : adj[pos]) {
			int cur = 0;

			// If our destination isn't blocked and it's 1 closer to the sink and there's flow, we
			// can go this way.
			if(!blocked[e.v2] && dist[e.v2] == dist[pos]-1 && e.cap - e.flow > 0) {

				// Recursively run dfs from here - limiting flow based on current and what's left on this edge.
				cur = dfs(e.v2, Math.min(min-flow, e.cap - e.flow));

				// Add the flow through this edge and subtract it from the reverse flow.
				e.flow += cur;
				e.rev.flow = -e.flow;

				// Add to the total flow.
				flow += cur;
			}

			// No more can go through, we're good.
			if(flow == min)
				return flow;
		}

		// mark if this node is now blocked.
		blocked[pos] = flow != min;

		// This is the flow
		return flow;
	}

	public int flow() {
		clear();
		int ret = 0;

		// Run a top level BFS.
		while(bfs()) {

			// Reset this.
			Arrays.fill(blocked, false);

			// Run multiple DFS's until there is no flow left to push through.
			ret += dfs(s, oo);
		}
		return ret;
	}

	// Just resets flow through all edges to be 0.
	public void clear() {
		for(ArrayList<Edge> edges : adj)
			for(Edge e : edges)
				e.flow = 0;
	}
}