#include <algorithm>
#include <cassert>
#include <cstring>
#include <iostream>
#include <map>

using namespace std;

typedef long long llint;

const int MAX = 300300;
const llint inf = 1e18;

int a[MAX];
map<int, llint> S[MAX]; // one-way
map<int, llint> P[MAX]; // round-trip

llint solve(string q, llint cA, llint cB, llint cAB, llint cBA) {
  int N = q.size();
  
  if (cAB > cBA) { // make cAB <= cBA
    swap(cAB, cBA);
    swap(cA, cB);
    for (auto& c: q) {
      c ^= 'A' ^ 'B';
    }
  }
  
  cA = min(cA, cAB);
  cB = min(cB, cBA);
  // now round-trip tickets must be used completely

  int cnt_A = count(q.begin(), q.end(), 'A');
  int cnt_B = N - cnt_A;

  // 3 cases:
  if (cA + cB <= cAB) {
    return cnt_A*cA + cnt_B*cB ;
  }

  int K = 0;
  int next_B = 0;
  for (int i = 0; i < N; ++i) {
    if (q[i] == 'A') {
      while (next_B < N && (next_B < i || q[next_B] != 'B')) next_B++;
      if (next_B < N) {
        K++;
        next_B++;
      }
    }
  }
  // K = max. number of A-B matches, where B is after A
  
  if (cAB <= cA + cB && cA + cB <= cBA) {
    return K*cAB + (cnt_A - K)*cA + (cnt_B - K)*cB;
  }

  int L = min(cnt_A, cnt_B) - K;
  // L = max. number of B-A matches, after K A-B matches are used
  
  if (cAB <= cA + cB && cBA <= cA + cB) {
    return K*cAB + L*cBA + (cnt_A - K - L)*cA + (cnt_B - K - L)*cB;
  }

  assert(false);
}

int main(void) {
  int N, D;
  scanf("%d %d", &N, &D);
  for (int i = 0; i < D; ++i) {
    scanf("%d", &a[i]);
    --a[i];
  }

  int M;
  scanf("%d", &M);
  for (int i = 0; i < M; ++i) {
    char tip[5];
    int a, b;
    llint x;
    scanf("%d %d %s %lld", &a, &b, tip, &x); --a, --b;

    if (tip[0] == 'O') {
      if (!S[a].count(b) || x < S[a][b]) S[a][b] = x;
    }
    if (tip[0] == 'R') {
      if (!P[a].count(b) || x < P[a][b]) P[a][b] = x;
    }
  }

  // group trips by city pairs
  map<pair<int, int>, string> T;
  for (int i = 0; i + 1 < D; ++i) {
    int x = a[i];
    int y = a[i + 1];
    if (x < y) T[{x, y}].push_back('A');
    else T[{y, x}].push_back('B');
  }

  llint ans = 0;
  for (auto& p: T) {
    int x = p.first.first;
    int y = p.first.second;
    string q = p.second;

    if (!S[x].count(y)) S[x][y] = inf;
    if (!P[x].count(y)) P[x][y] = inf;
    if (!S[y].count(x)) S[y][x] = inf;
    if (!P[y].count(x)) P[y][x] = inf;
    
    llint cost = solve(q, S[x][y], S[y][x], P[x][y], P[y][x]);
    ans += cost;
  }

  printf("%lld\n", ans);
  return 0;
}