#include<string.h>
#include<iostream>
#include<sstream>
#include<string>
#include<stdio.h>
#include<map>
using namespace std;
int l[2], n;
string v1[30][2], v2[30][2], v3[30][2];
char op[30][2];
int n1[30][2], n2[30][2], n3[30][2];
map<string, int>mp;
struct node
{
	double t[12];
	double r[2][2];
	double p;
	void init()
	{
		memset(t, 0, sizeof(t));
		memset(r, 0, sizeof(r));
		p = 1.0;
	}
	void run(int id, int o)
	{
		int V1, V2, V3;
		double u1, u2;
		V1 = mp[v1[(o - 1) / 4][id]];
		V2 = mp[v2[(o - 1) / 4][id]];
		V3 = mp[v3[(o - 1) / 4][id]];
		if (V1)
			u1 = t[V1];
		else
			u1 = n1[(o - 1) / 4][id];
		if (V2)
			u2 = t[V2];
		else
			u2 = n2[(o - 1) / 4][id];
		switch (o % 4)
		{
		case 1:r[0][id] = u1; break;
		case 2:r[1][id] = u2; break;
		case 3:
			if (op[(o - 1) / 4][id] == '+')
				r[0][id] = r[0][id] + r[1][id];
			else
				r[0][id] = r[0][id] - r[1][id]; break;
		case 0:t[V3] = r[0][id]; break;
		}

	}
}dp[125][125];
bool check(string s, int &a)//读取常量
{
	if (s[0] - '0' >= 0 && s[0] - '0' <= 9)
	{
		istringstream sin(s);
		sin >> a;
		return 0;
	}
	return 1;
}
void solve()
{
	int i, j, k, k1, k2;
	double p1, p2, p;
	node f1, f2;
	dp[0][0].init();
	for (i = 0; i <= l[0]; i++)
		for (j = 0; j <= l[1]; j++)
		{
			if (i | j)
			{
				if (i == 0)
				{
					f1 = dp[i][j - 1];
					p1 = f1.p*0.5; p2 = 0;
					f1.run(1, j);
				}
				else if (j == 0)
				{
					f2 = dp[i - 1][j];
					p2 = f2.p*0.5;
					p1 = 0;
					f2.run(0, i);
				}
				else
				{
					f1 = dp[i][j - 1];
					f2 = dp[i - 1][j];
					p1 = f1.p*(i == l[0] ? 1 : 0.5);
					p2 = f2.p*(j == l[1] ? 1 : 0.5);
					f1.run(1, j);
					f2.run(0, i);
				}
				dp[i][j].p = p = p1 + p2;
				for (k = 1; k <= n; k++)
				{
					dp[i][j].t[k] = (f1.t[k] * p1 + f2.t[k] * p2) / p;
				}

				for (k1 = 0; k1 <= 1; k1++)
					for (k2 = 0; k2 <= 1; k2++)
					{
						dp[i][j].r[k1][k2] = (f1.r[k1][k2] * p1 + f2.r[k1][k2] * p2) / p;
					}
			}
		}
	for (i = 1; i <= n; i++)
		if (dp[l[0]][l[1]].t[i] == 0)
			printf("0.0000\n");
		else
			printf("%.4lf\n", dp[l[0]][l[1]].t[i]);
}
int main()
{
	int ti, i, j;
	char ch;
	string str;
	scanf("%d", &ti);
	while (cin.peek() == '\n')
		getchar();
	while (ti--)
	{
		mp.clear();
		for (i = 0; i <= 1; i++)
		{
			while (cin.peek() == '\n')//吸收多余空行
				getchar();
			for (j = 0; 1; j++)
			{
				v1[j][i] = v2[j][i] = v3[j][i] = "";
				n1[j][i] = n2[j][i] = n3[j][i] = 0;
				while (cin.peek() != ':'&&cin.peek() != '\n')
				{
					ch = getchar();
					if (ch != ' ')
						v3[j][i] += toupper(ch);
				}
				if (v3[j][i] == "END")
					break;
				if (check(v3[j][i], n3[j][i]) == 1)
					mp[v3[j][i]]++;
				for (ch = getchar(); cin.peek() == ' ';)
					ch = getchar();
				for (ch = getchar(); cin.peek() == ' ';)
					ch = getchar();
				while (cin.peek() != '+'&&cin.peek() != '-')
				{
					ch = getchar();
					if (ch != ' ')
						v1[j][i] += toupper(ch);
				}
				if (check(v1[j][i], n1[j][i]) == 1)
					mp[v1[j][i]]++;
				scanf("%c", &op[j][i]);
				for (; cin.peek() == ' ';)
					ch = getchar();
				while (cin.peek() != ' '&&cin.peek() != '\n')
				{
					ch = getchar();
					if (ch != ' ')
						v2[j][i] += toupper(ch);
				}
				if (check(v2[j][i], n2[j][i]) == 1)
					mp[v2[j][i]]++;
				while (cin.peek() == ' ')
					ch = getchar();
				while (cin.peek() == '\n')
					getchar();
			}
			l[i] = j * 4;
		}
		map<string, int>::iterator it = mp.begin();//map内部已排序
		for (i = 1; it != mp.end(); i++, it++)
			mp[it->first] = i;
		n = i - 1;
		solve();
		if (ti!=0)
			printf("\n");
	}
}