// CERC 2012
// Problem E: Word Equations
// Brute force solution (forward traversal), O(tp)
// Author: Arkadiusz Pawlik

#include <cassert>
#include <iostream>
#include <map>
#include <string>
#include <vector>
using namespace std;

const int MAXK = 1000;
const int MAXS = 5;
const int MAXT = 5000;

struct eq
{
    string      s[2];
    eq *        e[2];
    
    eq(string s0="", string s1="")
    {
        s[0] = s0; s[1] = s1;
        e[0] = e[1] = 0;
    }

    int term(string const & p, int j)
    {
        int maxi = s[0].length();
        int maxj = p.length();
        for (int i=0; i<maxi && j<maxj; ++i)
            if (s[0][i] == p[j])
                ++j;
        return j;
    }

    int best(string const & p, int j)
    {
        return e[1] ? e[1]->best(p, e[0]->best(p, j)) : term(p, j);
    }
};

int main()
{
    int z; for (cin >> z; z; --z)
    {
        map<string, eq> M;
        string s[3], t, p;
        int k; for (cin >> k; k; --k)
        {
            cin >> s[2] >> s[1] >> s[0];
            if (s[0][0]<='Z') cin >> s[1] >> s[1];
            M[s[2]] = eq(s[0], s[1]);
        }
        for (map<string, eq>::iterator i = M.begin(); i != M.end(); ++i)
            if (i->second.s[0][0]<='Z')
                for (int k=0; k<2; ++k)
                    i->second.e[k] = &M[i->second.s[k]];
        cin >> t >> p;
        cout << (M[t].best(p, 0) < p.length() ? "NO" : "YES") << endl;
    }
    return 0;
}