/*
大模擬問題
*/
 
#include <bits/stdc++.h>
using namespace std;
 
//m：總?cè)藬?shù)  n：始終說謊人數(shù)  p：說話的總數(shù)
int m, n,  p;
//judge[i]：第i句話是真是假，真1假-1不清楚0  w[i]：第i局話是編號多少的的人說的
int judge[21], w[200];
//err：矛盾標記  nx：當前可能的罪犯
int err,  nx;
//name[i]：所有人名字（編號為1~m）  say[i]：所有人說的話  day[i]：所有星期幾名字
string name[100], say[200];
string day[10] = {"0", "Today is Sunday.", "Today is Monday.",
                  "Today is Tuesday.", "Today is Wednesday.", "Today is Thursday.",
                  "Today is Friday.", "Today is Saturday.",
                 };
 
//sset函數(shù)標記一個人說話真假
void sset(int who, int x) {
	if (judge[who] == -x)
		err = 1; //如果一個人既說真話又說假話，則矛盾
	else
		judge[who] = x;
}
 
int main() {
	cin >> m >> n >> p;
	for (int i = 1; i <= m; i++) {
		cin >> name[i];
	}
	for (int i = 1; i <= p; i++) {
		string nm;
		cin >> nm; //輸入說這句話人的名字
		nm.erase(nm.end() - 1); //刪除nm中冒號，便于判斷這句話編號多少的的人說的
		for (int j = 1; j <= m; j++) {
			if (name[j] == nm)
				w[i] = j;
		}
		getline(cin, say[i]);
		say[i].erase(say[i].begin()); //刪除say[i]中的起始空格
	}
 
	for (int td = 1; td <= 7; td++) { //暴力枚舉今天是星期幾
		for (int px = 1; px <= m; px++) { //暴力枚舉罪犯編號是幾號
			err = 0; //清除標記
			memset(judge, 0, sizeof(judge)); //初始化為不清楚真假
			//依次判斷每一句說的話
			for (int i = 1; i <= p; i++) {
				int who = w[i]; //說這句話人的編號
				//如果一個人是罪犯，并且說自己是罪犯，則說的就是真話，否則就是假話
				if (say[i] == "I am guilty.")
					sset(who, px == who ? 1 : -1);
				//如果一個人不是罪犯，并且說自己不是罪犯，則說的就是真話，否則就是假話
				if (say[i] == "I am not guilty.")
					sset(who, px != who ? 1 : -1);
				//如果一個人說今天是星期幾，說對了就是真話，說錯了就是假話
				for (int j = 1; j <= 7; j++) {
					if (say[i] == day[j])
						sset(who, j == td ? 1 : -1);
				}
				//如果一個人說其他人不是罪犯，說對了就是真話，說錯了就是假話
				for (int j = 1; j <= m; j++) {
					if (say[i] == name[j] + " is guilty.")
						sset(who, j == px ? 1 : -1);
					if (say[i] == name[j] + " is not guilty.")
						sset(who, j != px ? 1 : -1);
				}
			}
			int cnt = 0; //說假話的人數(shù)
			int no = 0; //不清楚真假的的人數(shù)
			for (int i = 1; i <= m; i++) {
				if (judge[i] == -1) //假
					cnt++;
				if (judge[i] == 0) //不清楚
					no++;
			}
			//如果出現(xiàn)Impossible的情況，err = 1，出現(xiàn)矛盾
			//如果cnt<=n<=cnt+no,即假設(shè)合理
			if (!err && cnt <= n && cnt + no >= n) {
				if (nx && nx != px) { //如果出現(xiàn)了兩個合理的罪犯
					cout << "Cannot Determine";
					return 0;
				} else {
					nx = px;
				}
			}
		}
	}
	if (!nx)
		cout << "Impossible";
	else
		cout << name[nx];
 
	return 0;
}