//判定是否存在从某点到0点的欧拉回路

//Time:0Ms	Memory:116K

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

#define MAX 25

int st, n;

int door[MAX];

int main()

{

	char s[120];

	while (scanf("%s", s), strcmp(s, "ENDOFINPUT"))

	{

		memset(door, 0, sizeof(door));

		int doors = 0;

		scanf("%d%d", &st, &n);

		gets_s(s, 120);

		for (int i = 0; i < n; i++)

		{

			gets_s(s, 120);

			int num, k = 0;

			while (sscanf(s + k, "%d", &num) == 1)

			{

				doors++;

				door[num]++;

				door[i]++;

				while (s[k] == ' ')	k++;

				while (s[k] && s[k] != ' ') k++;

			}

		}

		gets_s(s, 120);

		int odd = 0;

		for (int i = 0; i < n; i++)

			odd += door[i] % 2 == 1;

		if (odd == 0 && st == 0)

			printf("YES %d\n", doors);	//无奇度节点

		else if (odd == 2 && st && door[st] % 2 && door[0] % 2)

			printf("YES %d\n", doors);	//两个奇度节点

		else printf("NO\n");

	}

	return 0;

}

//判断能否使给定的词组前后接龙

//Time:344Ms	Memory:120K

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

#define MAX 28

#define MAXS 1005

#define MAXN 100005

int n;

char s[MAXS];

int in[MAX], out[MAX];

int fa[MAX];

int find(int x)

{

	return fa[x] < 0 ? x : find(fa[x]);

}

int Union(int r1, int r2)

{

	r1 = find(r1);	r2 = find(r2);

	if (r1 == r2)	return r1;

	int tmp = fa[r1] + fa[r2];

	if (fa[r1] > fa[r2])

	{

		fa[r1] = r2;

		fa[r2] = tmp;

		return r2;

	}

	else {

		fa[r2] = r1;

		fa[r1] = tmp;

		return r1;

	}

}

int main()

{

	//freopen("words.in", "r", stdin);

	int T;

	scanf("%d", &T);

	while (T--) {

		memset(in, 0, sizeof(in));

		memset(out, 0, sizeof(out));

		memset(fa, -1, sizeof(fa));

		int pa;

		scanf("%d", &n);

		while (n--) {

			scanf("%s", s);

			int i = s[strlen(s) - 1] - 'a';

			int o = s[0] - 'a';

			out[o]++;	in[i]++;

			pa = Union(i, o);

		}

		int odd = 0;

		bool connect = true;

		bool A = false, B = false;

		for (int i = 0; i < 26; i++)

		{

			if (!in[i] && !out[i]) continue;

			if (pa != find(i)) {

				connect = false;	break;

			}

			if (in[i] - out[i] != 0)

			{

				odd++;

				if (in[i] - out[i] == 1)	A = true;

				if (in[i] - out[i] == -1)	B = true;

			}

		}

		if (connect && ((odd == 2 && A && B) || odd == 0))

			printf("Ordering is possible.\n");

		else printf("The door cannot be opened.\n");

	}

	return 0;

}