#include <iostream>
#include <cstring>
#include <algorithm>

#define x first
#define y second

using namespace std;

typedef pair<int, pair<int, int>> PII;

const int N = 110;

bool st[30];
PII stk[N];

int get_num(char s[])
{
	int len = strlen(s);
	int tmp = 0;
	for (int i = 0; i < len; i ++ )
		tmp = tmp * 10 + s[i] - '0';
	return tmp;
}

int calc(int top)
{
	int t = 0;
	for (int i = 1; i <= top; i ++ )
	{
		int st = stk[i].y.x, ed = stk[i].y.y;
		if (st && ed && st > ed) return t;
		if (!st && ed) return t;
		if (st && !ed) t ++ ;
	}
	return t;
}

int main()
{
	int T;
	scanf("%d", &T);
	while (T -- )
	{
		int n;
		char s[100];
		scanf("%d%s", &n, s);
		int t = 0;
		if (s[2] == 'n' && s[3] == ')') t = 1;
		else if (s[2] == 'n')
		{
			int pos = 4;
			while (s[pos] != ')')
				t = t * 10 + s[pos] - '0', pos ++ ;
		}
		
		memset(st, 0, sizeof st);
		int top = 0;
		int res = 0;
		bool success = true;
		while (n -- )
		{
			char op[2];
			scanf("%s", op);
			if (*op == 'F')
			{
				char s1[5], s2[5], s3[5];
				scanf("%s%s%s", s1, s2, s3);
				if (st[s1[0] - 'a'])
				{
					success = false;
					break;
				}
				else
				{
					st[s1[0] - 'a'] = true;
					int a1 = 0, a2 = 0;
					if (s2[0] != 'n') a1 = get_num(s2);
					if (s3[0] != 'n') a2 = get_num(s3);
					stk[ ++ top] = {s1[0] - 'a', {a1, a2}};
					//cout << calc(top) << '\n';
					res = max(res, calc(top));
				}
			}
			else
			{
				if (!top)
				{
					success = false;
					break;
				}
				PII t = stk[top -- ];
				st[t.x] = false;
				res = max(res, calc(top));
			}
		}
		if (top) success = false;
		
		if (!success) puts("ERR");
		else if (res == t) puts("Yes");
		else puts("No");
	}
	return 0;
}

总是读入错描述时间复杂度的字符串，把 O(n^11) 读成 O(n^12)