You are given a string S and an array of strings [t_1, t_2, ..., t_k]. Each string t_i consists of lowercase Latin letters from a to n; S consists of lowercase Latin letters from a to n and no more than 14 question marks.

Each string t_i has its cost c_i — an integer number. The value of some string T is calculated as ∑_{i = 1}^{k} F(T, t_i) ⋅ c_i, where F(T, t_i) is the number of occurences of string t_i in T as a substring. For example, F(aaabaaa, aa) = 4.

You have to replace all question marks in S with pairwise distinct lowercase Latin letters from a to n so the value of S is maximum possible.

Input

The first line contains one integer k (1 ≤ k ≤ 1000) — the number of strings in the array [t_1, t_2, ..., t_k].

Then k lines follow, each containing one string t_i (consisting of lowercase Latin letters from a to n) and one integer c_i (1 ≤ |t_i| ≤ 1000, -10^6 ≤ c_i ≤ 10^6). The sum of lengths of all strings t_i does not exceed 1000.

The last line contains one string S (1 ≤ |S| ≤ 4 ⋅ 10^5) consisting of lowercase Latin letters from a to n and question marks. The number of question marks in S is not greater than 14.

Output

Print one integer — the maximum value of S after replacing all question marks with pairwise distinct lowercase Latin letters from a to n.

Examples

Input


4
abc -10
a 1
b 1
c 3
?b?


Output


5


Input


2
a 1
a 1
?


Output


2


Input


3
a 1
b 3
ab 4
ab


Output


8


Input


1
a -1
?????????????


Output


0


Input


1
a -1
??????????????


Output


-1
#include <bits/stdc++.h>
const int ACAM_SZ = 1005;
const int SIGMA_SZ = 14;
const int FRAG_CNT = 16;
int ch[ACAM_SZ][SIGMA_SZ], fail[ACAM_SZ];
long long val[ACAM_SZ];
int total = 1;
void insert(std::string& s, int v) {
  int x = 1;
  for (auto c : s) {
    if (!ch[x][c - 'a']) ch[x][c - 'a'] = ++total;
    x = ch[x][c - 'a'];
  }
  val[x] += v;
}
void build_acam() {
  std::queue<int> Q;
  std::fill(ch[0], ch[0] + SIGMA_SZ, 1);
  Q.push(1);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int c = 0; c < SIGMA_SZ; c++) {
      if (!ch[x][c]) {
        ch[x][c] = ch[fail[x]][c];
        continue;
      }
      fail[ch[x][c]] = ch[fail[x]][c], Q.push(ch[x][c]);
      val[ch[x][c]] += val[fail[ch[x][c]]];
    }
  }
}
long long run(int& x, std::string s) {
  long long ret = 0;
  for (auto c : s) ret += val[x = ch[x][c - 'a']];
  return ret;
}
int n;
int l[FRAG_CNT], r[FRAG_CNT], fr_cnt;
long long ans;
std::string str;
int trans[ACAM_SZ][FRAG_CNT];
long long accmu[ACAM_SZ][FRAG_CNT];
long long dp[ACAM_SZ][1 << SIGMA_SZ];
signed main() {
  std::cin >> n;
  for (int i = 1, v; i <= n; i++) std::cin >> str >> v, insert(str, v);
  build_acam();
  std::cin >> str;
  for (int i = 0; i <= (int)str.size(); i++) {
    static std::string cur("");
    if (i == (int)str.size() || str[i] == '?') {
      for (int x = 1; x <= total; x++)
        accmu[x][fr_cnt] = run(trans[x][fr_cnt] = x, cur);
      cur.clear(), ++fr_cnt;
    } else
      cur += str[i];
  }
  memset(dp, -0x3f, sizeof(dp)), ans = dp[0][0];
  dp[trans[1][0]][0] = accmu[1][0];
  for (int S = 0; S < (1 << SIGMA_SZ); S++) {
    int cnt = __builtin_popcount(S) + 1;
    if (cnt > fr_cnt) continue;
    if (cnt == fr_cnt) {
      for (int i = 1; i <= total; i++) ans = std::max(ans, dp[i][S]);
      continue;
    }
    for (int i = 1; i <= total; i++)
      if (dp[i][S] > -1e16) {
        for (int c = 0; c < SIGMA_SZ; c++)
          if (!(S >> c & 1)) {
            int j = trans[ch[i][c]][cnt], T = S | (1 << c);
            dp[j][T] = std::max(
                dp[j][T], dp[i][S] + val[ch[i][c]] + accmu[ch[i][c]][cnt]);
          }
      }
  }
  std::cout << ans << std::endl;
  return 0;
}
