有N个人，各自有一个姓名和ID（别名）。每个人的姓名和ID都没有重复。这些人依次进入一间房间，然后可能会离开。过程中可以得到一些信息，告知在房间里的某个人的ID。你的任务是准确地确定每个人的ID。

共N行，每行形如：“姓名:ID”，如果ID不能确定，输出???。

按照姓名的字典顺序输出。

7
bigman mangler sinbad fatman bigcheese frenchie capodicapo
E mugsy
E knuckles
M bigman
M mangler
L mugsy
E clyde
E bonnie
M bigman
M fatman
M frenchie
L clyde
M fatman
E ugati
M sinbad
E moriarty
E booth
Q

bonnie:fatman
booth:???
clyde:frenchie
knuckles:bigman
moriarty:???
mugsy:mangler
ugati:sinbad

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <string>
#include <map>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
	return x * f;
}

#define maxn 45
#define maxm 3210
#define oo 2147483647
struct Edge {
	int from, to, flow;
	Edge() {}
	Edge(int _1, int _2, int _3): from(_1), to(_2), flow(_3) {}
} ;
struct Dinic {
	int n, m, s, t, head[maxn], next[maxm];
	Edge es[maxm];
	int hd, tl, Q[maxn], vis[maxn];
	int cur[maxn];
	void cpy(Dinic& T) {
		n = T.n; m = T.m; s = T.s; t = T.t;
		memcpy(head, T.head, sizeof(head));
		memcpy(next, T.next, sizeof(next));
		for(int i = 0; i < m; i++) es[i] = T.es[i];
		return ;
	}
	void init(int nn) {
		n = nn; m = 0;
		memset(head, -1, sizeof(head));
		return ;
	}
	void AddEdge(int a, int b, int c) {
		es[m] = Edge(a, b, c); next[m] = head[a]; head[a] = m++;
		es[m] = Edge(b, a, 0); next[m] = head[b]; head[b] = m++;
		return ;
	}
	bool BFS() {
		memset(vis, 0, sizeof(vis)); vis[s] = 1;
		hd = tl = 0; Q[++tl] = s;
		while(hd < tl) {
			int u = Q[++hd];
			for(int i = head[u]; i != -1; i = next[i]) {
				Edge& e = es[i];
				if(!vis[e.to] && e.flow) {
					vis[e.to] = vis[u] + 1;
					Q[++tl] = e.to;
				}
			}
		}
		return vis[t] > 1;
	}
	int DFS(int u, int a) {
		if(u == t || !a) return a;
		int flow = 0, f;
		for(int& i = cur[u]; i != -1; i = next[i]) {
			Edge& e = es[i];
			if(vis[e.to] == vis[u] + 1 && (f = DFS(e.to, min(a, e.flow)))) {
				flow += f; a -= f;
				e.flow -= f; es[i^1].flow += f;
				if(!a) return flow;
			}
		}
		return flow;
	}
	int MaxFlow(int ss, int tt) {
		s = ss; t = tt;
		int flow = 0;
		while(BFS()) {
			for(int i = 1; i <= n; i++) cur[i] = head[i];
			flow += DFS(s, oo);
		}
		return flow;
	}
} sol, sol2, sol3;

#define maxp 25
string ID[maxp], name[maxp];
map <string, int> M, M2;
bool S[maxp];
int en[maxn][maxn];
struct Info {
	string name, ID;
	Info() {}
	Info(string _, string __): name(_), ID(__) {}
	bool operator < (const Info& t) const { return name < t.name; }
} is[maxp];

int main() {
	int n = read();
	sol.init((n << 1) + 2); int s = (n << 1) + 1, t = s + 1;
	for(int i = 1; i <= n; i++) cin >> ID[i];
	sort(ID + 1, ID + n + 1);
	for(int i = 1; i <= n; i++) M[ID[i]] = i;
	int cp = 0;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++) en[i][j+n] = en[j+n][i] = 1;
	while(1) {
		char tc[2]; scanf("%s", tc);
		if(tc[0] == ‘Q‘) break;
		string t; cin >> t;
		if(tc[0] == ‘E‘) {
			if(!M2.count(t)) M2[t] = ++cp, name[cp] = t;
			S[M2[t]] = 1;
		}
		if(tc[0] == ‘L‘) S[M2[t]] = 0;
		if(tc[0] == ‘M‘) {
			int p = M[t];
			for(int i = 1; i <= n; i++) if(!S[i]) en[i][p+n] = en[p+n][i] = -1;
		}
	}
	int ce = 0;
	for(int i = 1; i <= cp; i++)
		for(int j = n + 1; j <= (n << 1); j++) if(en[i][j] > 0)
			sol.AddEdge(i, j, 1), en[i][j] = ce, ce += 2;
	for(int i = 1; i <= cp; i++) sol.AddEdge(s, i, 1);
	for(int i = 1; i <= n; i++) sol.AddEdge(i + n, t, 1);
	sol3.cpy(sol); sol2.cpy(sol);
	int ans = sol.MaxFlow(s, t);
	for(int i = 1; i <= cp; i++) {
		for(int j = n + 1; j <= (n << 1); j++) if(en[i][j] >= 0 && !sol.es[en[i][j]].flow) {
			sol2.cpy(sol3);
			sol2.es[en[i][j]].flow = 0;
			int tmp = sol2.MaxFlow(s, t);
//			printf("tmp: %d\n", tmp);
			if(tmp == ans) is[i] = Info(name[i], string("???"));
			else is[i] = Info(name[i], ID[j-n]);
			break;
		}
	}
	
	sort(is + 1, is + cp + 1);
	for(int i = 1; i <= cp; i++) cout << is[i].name << ‘:‘ << is[i].ID << endl;
	
	return 0;
}