D. Fedor and Essaytime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAfter you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the English teacher told him to prepare an essay. Fedor didn't want to prepare the essay, so he asked Alex for help. Alex came to help and wrote the essay for Fedor. But Fedor didn't like the essay at all. Now Fedor is going to change the essay using the synonym dictionary of the English language.
Fedor does not want to change the meaning of the essay. So the only change he would do: change a word from essay to one of its synonyms, basing on a replacement rule from the dictionary. Fedor may perform this operation any number of times.
As a result, Fedor wants to get an essay which contains as little letters «R» (the case doesn't matter) as possible. If there are multiple essays with minimum number of «R»s he wants to get the one with minimum length (length of essay is the sum of the lengths of all the words in it). Help Fedor get the required essay.
Please note that in this problem the case of letters doesn't matter. For example, if the synonym dictionary says that word cat can be replaced with word DOG, then it is allowed to replace the word Cat with the word doG.
InputThe first line contains a single integer m (1 ≤ m ≤ 105) — the number of words in the initial essay. The second line contains words of the essay. The words are separated by a single space. It is guaranteed that the total length of the words won't exceed 105 characters.
The next line contains a single integer n (0 ≤ n ≤ 105) — the number of pairs of words in synonym dictionary. The i-th of the next nlines contains two space-separated non-empty words xi and yi. They mean that word xi can be replaced with word yi (but not vise versa). It is guaranteed that the total length of all pairs of synonyms doesn't exceed 5·105 characters.
All the words at input can only consist of uppercase and lowercase letters of the English alphabet.
OutputPrint two integers — the minimum number of letters «R» in an optimal essay and the minimum length of an optimal essay.
Sample test(s)input3
AbRb r Zz
4
xR abRb
aA xr
zz Z
xr youtput2 6input2
RuruRu fedya
1
ruruRU fedoroutput1 10
1 /*************************************************************************
2 > File Name: D.cpp
3 > Author: Stomach_ache
4 > Mail: sudaweitong@gmail.com
5 > Created Time: 2014年09月19日 星期五 14时41分44秒
6 > Propose:
7 ************************************************************************/
8 #include <map>
9 #include <cmath>
10 #include <string>
11 #include <cstdio>
12 #include <vector>
13 #include <fstream>
14 #include <cstring>
15 #include <iostream>
16 #include <algorithm>
17 using namespace std;
18 /*Let's fight!!!*/
19
20 const int MAX_N = 100050;
21 const int INF = 0x3f3f3f3f;
22 typedef pair<int, int> pii;
23 typedef long long LL;
24
25 int V; // 顶点数
26 vector<int> G[MAX_N], rG[MAX_N], vs;
27 bool used[MAX_N];
28 int cmp[MAX_N];
29 //题目变量
30 map<string, int> HASH;
31 int n, m, ID[MAX_N], X[MAX_N], Y[MAX_N];
32 pii s[MAX_N], ss[MAX_N], dp[MAX_N];
33
34 void add_edge(int from, int to) {
35 G[from].push_back(to);
36 rG[to].push_back(from);
37 }
38
39 void dfs(int v) {
40 used[v] = true;
41 for (int i = 0; i < G[v].size(); i++) {
42 if (!used[G[v][i]]) dfs(G[v][i]);
43 }
44 vs.push_back(v);
45 }
46
47 void rdfs(int v, int k) {
48 used[v] = true;
49 cmp[v] = k;
50 ss[k] = min(ss[k], s[v]);
51 for (int i = 0; i < rG[v].size(); i++) {
52 if (!used[rG[v][i]]) rdfs(rG[v][i], k);
53 }
54 }
55
56 int scc() {
57 memset(used, false, sizeof(used));
58 vs.clear();
59 for (int v = 1; v <= V; v++) {
60 if (!used[v]) dfs(v);
61 }
62 memset(used, false, sizeof(used));
63 int k = 0;
64 for (int i = vs.size() - 1; i >= 0; i--) {
65 if (!used[vs[i]]) {
66 ss[++k] = pii(INF, INF);
67 rdfs(vs[i], k);
68 }
69 }
70 return k;
71 }
72
73
74 int get(string &str) {
75 for (int i = 0; i < str.size(); i++) {
76 str[i] = tolower(str[i]);
77 }
78 if (HASH.find(str) == HASH.end()) {
79 HASH[str] = ++V;
80 s[V].second= str.size();
81 for (int j = 0; j < s[V].second; j++) if (str[j] == 'r') s[V].first++;
82 return V;
83 } else {
84 return HASH[str];
85 }
86 }
87
88 void rebuild() {
89 for (int i = 1; i <= V; i++) G[i].clear();
90 for (int i = 1; i <= m; i++) if (cmp[X[i]] != cmp[Y[i]])
91 add_edge(cmp[X[i]], cmp[Y[i]]);
92 }
93
94 pii DFS(int u) {
95 if (used[u]) return dp[u];
96 used[u] = true;
97 dp[u] = ss[u];
98 for (int i = 0; i < G[u].size(); i++) {
99 dp[u] = min(dp[u], DFS(G[u][i]));
100 }
101 return dp[u];
102 }
103
104 int main(void) {
105 ios::sync_with_stdio(false);
106 cin >> n;
107 for (int i = 1; i <= n; i++) {
108 string str;
109 cin >> str;
110 int id = get(str);
111 ID[i] = id;
112 }
113
114 cin >> m;
115 for (int i = 1; i <= m; i++) {
116 string x, y;
117 cin >> x >> y;
118 int u = get(x), v = get(y);
119 add_edge(u, v);
120 X[i] = u, Y[i] = v;
121 }
122
123 int k = scc();
124 rebuild();
125
126 memset(used, false, sizeof(used));
127 LL resr = 0, resl = 0;
128 for (int i = 1; i <= n; i++) {
129 int pos = cmp[ID[i]];
130 DFS(pos);
131 resr += dp[pos].first;
132 resl += dp[pos].second;
133 }
134
135 cout << resr << ' ' << resl << endl;
136 return 0;
137 }