#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=double;
#define FOR(i,a,b) for(ll i=a;i<(ll)b;++i)
#define F(n) FOR(i,0,n)
#define FF(n) FOR(j,0,n)
#define aa first
#define bb second
#define PB push_back
#define EQ(a,b) (fabs(a-b)<=(fabs(a+b)*EPS))
#define MOD ((ll)(1e9+7))
#define INF (1ll << 60)

ll n; // n = p^k, size of input string
ll p;
string a;
ll prices[255]; // prices for each letter

// selected, size, offset
map<pair<ll,ll>,ll> noBest; // cycle not selected
map<pair<ll,ll>,ll> yesBest; // cycle selected
// used for temporary copy
map<pair<ll,ll>,ll> noBest2;
map<pair<ll,ll>,ll> yesBest2;

char cycleLetter(ll size, ll offset) {
  char found = '?';
  for (ll i = 0; i < size; i ++) {
    const char curChar = a[(i*n/size + offset)%n];
    if (curChar == '?') continue;
    else if (found == '?') found = curChar;
    //else if (found != curChar) return 0;
    else if (found != curChar) found = 0;
  }
  return found;
}

int main(){
  ios::sync_with_stdio(0);cin.tie(0);
  cin >> n >> a;
  p = 1;
  for (ll i = 2; i <= n; ++ i) {
    if (n%i == 0) {
      p = i;
      break;
    }
  }
  ll alpha;
  cin >> alpha;
  F(alpha) {
    char c;
    cin >> c >> prices[c];
    prices['?'] = max(prices['?'], prices[c]);
  }

  ll level = 0;
  for (ll size = 1; size <= n && (p != 1 || level == 0); size *= p) {
    level ++;
    for (ll offset = 0; offset < n/size; ++ offset) {
      // DONT SELECT
      if (size == 1) { // the trivial case
        noBest2[{size, offset}] = 0;
      } else {
        ll value = 0;
        for (ll i = 0; i < p; ++ i) {
          value += max(noBest[{size/p, i*n/size + offset}], yesBest[{size/p, i*n/size + offset}]);
        }
        noBest2[{size, offset}] = value;
      }

      // SELECT
      char letter = cycleLetter(size, offset);
      yesBest2[{size, offset}] = level * prices[letter] * size;
    }
    // keep the maps small
    yesBest = yesBest2;
    noBest = noBest2;
    yesBest2 = map<pair<ll,ll>,ll>();
    noBest2 = map<pair<ll,ll>,ll>();
  }

  cout << max(noBest[{n, 0}], yesBest[{n, 0}]) << endl;

  return 0;
}