//ti表示给定的复杂度最多有几层O(n)的循环
//now是现在的层数,maxn是记录的最大层数，check[]记录变量是否使用和对应复杂度(1是O(1),2是O(n))
//firstc记录第一个无法进入的循环的变量 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<cmath>
#include<stack>
#include<map>
#define ll long long
using namespace std;
int T,n,ti=0,now,maxn,sum=0;
int check[26];
stack<int>s;
char firstc='0';
bool flag=0;
void krual()//初始化 
{
	sum=0;
	firstc='0';
	flag=0;
	ti=0;
	maxn=0;
	now=0;
	memset(check,0,sizeof(check));
	return ;
}
int main()
{
	//freopen("temp.out","w",stdout);
	scanf("%d",&T);
	while(T--)
	{
		krual();
		
		scanf("%lld",&n);
		string s1,s2,s3,s4;
		char c;
		cin>>s1;
		if(s1[2] == 'n')
		{
			int len=s1.size();
			for(int i=4;i<len-1;++i) ti=ti*10+(s1[i]-'0');
		}
		while(!s.empty()) s.pop();
		while(n--)
		{
			//cout<<firstc<<" "<<now<<endl;
			cin>>c;
			if(c == 'F')
			{
				++sum;
				cin>>c>>s3>>s4;
				if(flag) continue;
				if(check[c-'a'] > 0)
				{
					flag=1;
					continue;
				}
				if(s4 == "n" && s3 !="n")//合法O(n) 
				{
					++now;
					if(firstc == '0')
					  maxn=max(maxn,now);
					s.push(c-'a');
					check[c-'a']=2;
				}
				else if((s3 == "n"  && s4 == "n" )|| (s3 != "n" && s4 != "n" && s3 <= s4))//合法O(1) 
				{
					s.push(c-'a');
					check[c-'a']=1;
				}
				else//非法 
				{
					if(firstc == '0') firstc=c;
					s.push(c-'a');
					check[c-'a']=1;
				}
			}
			else
			{
				--sum;
				if(s.empty()) flag=1;
				if(flag) continue;
				if(check[s.top()] == 2) --now;
				if(s.top() == (int)(firstc-'a')) firstc='0';
				check[s.top()]=0;
				s.pop();
			}
		}
		if(sum != 0) flag=1;
		if(flag) printf("ERR\n");
		else if(maxn != ti) printf("No\n");
		else printf("Yes\n");
	}
	return 0;
}