憨批题，怎么我CF的时候就做不出这个题啊啊啊啊啊啊啊
题意:
给你 (k)，(k)个串 (t_i) 以及价值 (c_i)((sum |t_i| leq 10^3))，然后给你一个串 (s)((|s|leq 4 imes 10^5))，(s) 中最多包含 (14) 个 (?)，字符集是 (14)，每个 (?) 不能取到相同字符，对于每一种填充字符的方式，求最大的(sum F(s,t_i) imes c_i)
sol:
ACAutoMaton 傻子题。
考虑到 AC 自动机如果把 (ed_{fa_i}) 信息加到 (ed_i) 上，那么 (ed_i) 就是 (i) 节点对应的字符串包含的所有子串的信息。
我们考虑到最多有 (14) 个 (?)，所以这个玩意最多被分成了 (14) 段。那么显然我们可以直接计算对于每个节点 (j) 在第 (i) 段会变成 AC 自动机上的哪个节点 (nxt_{j,i})，以及计算(j) 走过第 (i) 段的贡献 (sum_{j,i} = sum ed_p)，然后就可以 (dp) 了。
// clang-format off
// powered by c++11
// by Isaunoya
#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define Rep(i,x,y) for(register int i=(x);i>=(y);--i)
using namespace std;using db=double;using ll=long long;
using uint=unsigned int;using ull=unsigned long long;
using pii=pair<int,int>;
#define Tp template
#define fir first
#define sec second
Tp<class T>void cmax(T&x,const T&y){if(x<y)x=y;}Tp<class T>void cmin(T&x,const T&y){if(x>y)x=y;}
#define all(v) v.begin(),v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T>void sort(vector<T>&v){sort(all(v));}Tp<class T>void reverse(vector<T>&v){reverse(all(v));}
Tp<class T>void unique(vector<T>&v){sort(all(v)),v.erase(unique(all(v)),v.end());}inline void reverse(string&s){reverse(s.begin(),s.end());}
const int SZ=1<<23|233;
struct FILEIN{char qwq[SZ],*S=qwq,*T=qwq,ch;
#ifdef __WIN64
#define GETC getchar
#else
inline char GETC(){return(S==T)&&(T=(S=qwq)+fread(qwq,1,SZ,stdin),S==T)?EOF:*S++;}
#endif
inline FILEIN&operator>>(char&c){while(isspace(c=GETC()));return*this;}inline FILEIN&operator>>(string&s){s.clear();while(isspace(ch=GETC()));if(!~ch)return*this;s=ch;while(!isspace(ch=GETC())&&~ch)s+=ch;return*this;}
inline FILEIN&operator>>(char*str){char*cur=str;while(*cur)*cur++=0;cur=str;while(isspace(ch=GETC()));if(!~ch)return*this;*cur=ch;while(!isspace(ch=GETC())&&~ch)*++cur=ch;*++cur=0;return*this;}
Tp<class T>inline void read(T&x){bool f=0;while((ch=GETC())<48&&~ch)f^=(ch==45);x=~ch?(ch^48):0;while((ch=GETC())>47)x=x*10+(ch^48);x=f?-x:x;}
inline FILEIN&operator>>(int&x){return read(x),*this;}inline FILEIN&operator>>(ll&x){return read(x),*this;}inline FILEIN&operator>>(uint&x){return read(x),*this;}inline FILEIN&operator>>(ull&x){return read(x),*this;}
inline FILEIN&operator>>(double&x){read(x);bool f=x<0;x=f?-x:x;if(ch^'.')return*this;double d=0.1;while((ch=GETC())>47)x+=d*(ch^48),d*=.1;return x=f?-x:x,*this;}
}in;
struct FILEOUT{const static int LIMIT=1<<22;char quq[SZ],ST[233];int sz,O,pw[233];
FILEOUT(){set(7);rep(i,pw[0]=1,9)pw[i]=pw[i-1]*10;}~FILEOUT(){flush();}
inline void flush(){fwrite(quq,1,O,stdout),fflush(stdout),O=0;}
inline FILEOUT&operator<<(char c){return quq[O++]=c,*this;}inline FILEOUT&operator<<(string str){if(O>LIMIT)flush();for(char c:str)quq[O++]=c;return*this;}
inline FILEOUT&operator<<(char*str){if(O>LIMIT)flush();char*cur=str;while(*cur)quq[O++]=(*cur++);return*this;}
Tp<class T>void write(T x){if(O>LIMIT)flush();if(x<0){quq[O++]=45;x=-x;}do{ST[++sz]=x%10^48;x/=10;}while(x);while(sz)quq[O++]=ST[sz--];}
inline FILEOUT&operator<<(int x){return write(x),*this;}inline FILEOUT&operator<<(ll x){return write(x),*this;}inline FILEOUT&operator<<(uint x){return write(x),*this;}inline FILEOUT&operator<<(ull x){return write(x),*this;}
int len,lft,rig;void set(int l){len=l;}inline FILEOUT&operator<<(double x){bool f=x<0;x=f?-x:x,lft=x,rig=1.*(x-lft)*pw[len];return write(f?-lft:lft),quq[O++]='.',write(rig),*this;}
}out;
#define int long long
struct Math{
vector<int>fac,inv;int mod;
void set(int n,int Mod){fac.resize(n+1),inv.resize(n+1),mod=Mod;rep(i,fac[0]=1,n)fac[i]=fac[i-1]*i%mod;inv[n]=qpow(fac[n],mod-2);Rep(i,n-1,0)inv[i]=inv[i+1]*(i+1)%mod;}
int qpow(int x,int y){int ans=1;for(;y;y>>=1,x=x*x%mod)if(y&1)ans=ans*x%mod;return ans;}int C(int n,int m){if(n<0||m<0||n<m)return 0;return fac[n]*inv[m]%mod*inv[n-m]%mod;}
int gcd(int x,int y){return!y?x:gcd(y,x%y);}int lcm(int x,int y){return x*y/gcd(x,y);}
}math;
// clang-format on
const int maxn = 1e3 + 31;
const int maxm = 4e5 + 54;
char t[maxn];
char s[maxm];
int ans = -1e18;
struct acautomaton {
int cnt;
int ch[maxn][14], fail[maxn], ed[maxn];
acautomaton() { cnt = 1; }
void ins(char* s, int val) {
char* cur = s;
int p = 1;
while (*cur) {
int c = (*cur++) - 'a';
if (!ch[p][c]) ch[p][c] = ++cnt;
p = ch[p][c];
}
ed[p] += val;
}
void build() {
queue<int> q;
rep(i, 0, 13) if (ch[1][i]) fail[ch[1][i]] = 1, q.push(ch[1][i]);
else ch[1][i] = 1;
while (q.size()) {
int u = q.front();
q.pop();
rep(i, 0, 13) {
if (ch[u][i]) {
fail[ch[u][i]] = ch[fail[u]][i];
ed[ch[u][i]] += ed[fail[ch[u][i]]];
q.push(ch[u][i]);
} else {
ch[u][i] = ch[fail[u]][i];
}
}
}
}
int pos[16], tot = 0;
int nxt[maxn][16], sum[maxn][16];
int dp[maxn][1 << 14 | 233];
void solve(char* s) {
int len = strlen(s);
for (int i = 0; i < len; i++)
if (s[i] == '?') pos[++tot] = i;
pos[0] = -1;
pos[tot + 1] = len;
rep(i, 0, tot) {
rep(j, 1, cnt) {
int p = j;
rep(k, pos[i] + 1, pos[i + 1] - 1) {
p = ch[p][s[k] - 'a'];
sum[j][i] += ed[p];
}
nxt[j][i] = p;
}
}
rep(i, 1, cnt) rep(j, 0, (1 << 14)) dp[i][j] = -1e18;
dp[nxt[1][0]][0] = sum[1][0];
auto count = [&](int x) {
int c = 0;
while (x) x ^= x & -x, c++;
return c;
};
rep(i, 0, (1 << 14) - 1) {
int v = count(i);
if (v >= tot) continue;
rep(j, 1, cnt) {
rep(k, 0, 13) {
if (i & (1 << k)) continue;
cmax(dp[nxt[ch[j][k]][v + 1]][i ^ (1 << k)], dp[j][i] + sum[ch[j][k]][v + 1] + ed[ch[j][k]]);
}
}
}
rep(i, 0, (1 << 14) - 1) {
if (count(i) ^ tot) continue;
rep(j, 1, cnt) cmax(ans, dp[j][i]);
}
}
} acam;
signed main() {
// code begin.
int _;
in >> _;
while (_--) {
int val;
in >> t >> val, acam.ins(t, val);
}
acam.build(), in >> s, acam.solve(s);
out << ans << '
';
return 0;
// code end.
}