#include<bits/stdc++.h>
using namespace std;
typedef long long ll ;
int n;
ll val[26];
string s;
int k,p;
ll f[18][100005];
int fpow(int a,int b)
{
    int ans = 1;
    while(b--) ans *= a;
    return ans;
}
int main()
{
   // freopen("in.txt","r",stdin) ;
    ios::sync_with_stdio(false) ; cin.tie(0) ;cout.tie(0) ;
    cin >> n; cin >> s; cin >> k;
    for(int i = 0;i < 26;i++) val[i] = -1e18;
    for(int i = 1;i <= k;i++) {
        char c;
        int d;cin >> c >> d;
        val[c - 'A'] = d;
    }
    for(int i = 2;i <= n;i++) {
        int hs = 1,j;
        for(j = 0; hs < n;j++) {
            hs = hs * i;
        }
        if(hs == n) {
            p = i , k = j ; break ;
        }
    }
    int mx = 0;
    for(int i = 0;i < 26;i++) mx = max(1LL*mx , val[i]) ; 
    for(int i = k;i >= 0;i--) { ///步长为p^i
        int d = fpow(p , k - i) ; ///一圈的大小
        int step = fpow(p , i) ;
        for(int j = 0;j < step;j++) {
            int cha = -1;
            for(int k = j ; ; k = (k + step) % n) {
                if(s[k] != '?') {
                    if(cha == -1) cha = s[k] - 'A';
                    else {
                        if(cha != s[k] - 'A') {cha = -2;break ;}
                    }
                }
                if((k + step) % n == j) break ;
            }
            f[i][j] = -1e18;
            if(cha != -2) {
                if(cha == -1) f[i][j] = 1LL * mx * d * (k - i+1);
                else f[i][j] = 1LL*val[cha] * d * (k - i+1);
            }
            if(i != k) {
                ll f2 = 0;
                for(int u = 0;u < p;u++) f2 += f[i + 1][(j + u*step) % (step * p)] ;
                f[i][j] = max(f[i][j] , f2) ;
            }
            // printf("%d %d %lld\n",i,j,f[i][j]) ;
        }
    }
    cout << f[0][0] << '\n' ;
    return 0;
}