Problem: Mahmoud wants to send a message to his friend Ehab. Their language consists of n words numbered from 1 to n. Some words have the same meaning so there are k groups of words such that all the words in some group have the same meaning.

Mahmoud knows that the i-th word can be sent with cost ai. For each word in his message, Mahmoud can either replace it with another word of the same meaning or leave it as it is. Can you help Mahmoud determine the minimum cost of sending the message?

The cost of sending the message is the sum of the costs of sending every word in it.

Input

The first line of input contains integers n, k and m (1 ≤ k ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of words in their language, the number of groups of words, and the number of words in Mahmoud's message respectively.

The second line contains n strings consisting of lowercase English letters of length not exceeding 20 which represent the words. It's guaranteed that the words are distinct.

The third line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) where ai is the cost of sending the i-th word.

The next k lines describe the groups of words of same meaning. The next k lines each start with an integer x (1 ≤ x ≤ n) which means that there are x words in this group, followed by x integers which represent the indices of words in this group. It's guaranteed that each word appears in exactly one group.

The next line contains m space-separated words which represent Mahmoud's message. Each of these words appears in the list of language's words.

Output

The only line should contain the minimum cost to send the message after replacing some words (maybe none) with some words of the same meaning.

Examples

Input

5 4 4
i loser am the second
100 1 1 5 10
1 1
1 3
2 2 5
1 4
i am the second


Output

107

Input

5 4 4
i loser am the second
100 20 1 5 10
1 1
1 3
2 2 5
1 4
i am the second


Output

116

Note

In the first sample, Mahmoud should replace the word "second" with the word "loser" because it has less cost so the cost will be 100+1+5+1=107.

In the second sample, Mahmoud shouldn't do any replacement so the cost will be 100+1+5+10=116.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int gcd(long long a, int b) {
  if (a < b) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
class Triplet {
 public:
  long long gcd;
  long long x;
  long long y;
};
Triplet gcdExtendedEuclid(int a, int b) {
  if (b == 0) {
    Triplet myAns;
    myAns.gcd = a;
    myAns.x = 1;
    myAns.y = 0;
    return myAns;
  }
  Triplet smallAns = gcdExtendedEuclid(b, a % b);
  Triplet myAns;
  myAns.gcd = smallAns.gcd;
  myAns.x = smallAns.y;
  myAns.y = (smallAns.x - ((a / b) * (smallAns.y)));
  return myAns;
}
long long modInverse(int A, int M) {
  Triplet ans = gcdExtendedEuclid(A, M);
  return ans.x;
}
bool check(vector<long long>& v1, long long val, long long h) {
  long long ans = 0;
  for (long long i = 0; i < v1.size(); i++) {
    ans += v1[i] / val;
    if (v1[i] % val != 0) {
      ans += 1;
    }
  }
  if (ans > h) {
    return false;
  }
  return true;
}
void multiply(long long A[2][2], long long M[2][2]) {
  long long firstValue = (A[0][0] * M[0][0] + A[0][1] * M[1][0]) % 1000000007;
  long long secondValue = (A[0][0] * M[0][1] + A[0][1] * M[1][1]) % 1000000007;
  long long thirdValue = (A[1][0] * M[0][0] + A[1][1] * M[1][0]) % 1000000007;
  long long fourthValue = (A[1][0] * M[0][1] + A[1][1] * M[1][1]) % 1000000007;
  A[0][0] = firstValue;
  A[0][1] = secondValue;
  A[1][0] = thirdValue;
  A[1][1] = fourthValue;
  return;
}
void getpow(long long A[2][2], long long n) {
  if (n == 0 || n == 1) {
    return;
  }
  getpow(A, n / 2);
  multiply(A, A);
  if (n % 2 != 0) {
    long long first[2][2] = {{1, 1}, {1, 0}};
    multiply(A, first);
  }
  return;
}
long long getpow1(long long n, long long m, long long b) {
  if (m == 0) {
    return 1;
  }
  long long ans = getpow1(n, m / 2, b) % b;
  ans = (ans * ans) % b;
  if (m % 2 != 0) {
    ans = ans * n % b;
  }
  return ans % b;
}
long long getfib(long long n) {
  if (n == 0) {
    return 0;
  }
  long long A[2][2] = {{1, 1}, {1, 0}};
  getpow(A, n);
  long long first[1][2];
  first[0][0] = 0;
  first[0][1] = 1;
  long long ans = first[0][0] * A[0][0] + first[0][1] * A[1][0];
  return ans % 1000000007;
}
vector<long long>* sieve() {
  bool isprime[1000001];
  for (long long i = 0; i < 1000001; i++) {
    isprime[i] = true;
  }
  vector<long long>* primes = new vector<long long>();
  for (long long i = 2; i < 1000001; i++) {
    if (isprime[i] == true) {
      for (long long j = 2 * i; j < 1000001; j += i) {
        isprime[j] = false;
      }
    }
  }
  primes->push_back(2);
  for (long long i = 3; i < 1000001; i++) {
    if (isprime[i] == true) {
      primes->push_back(i);
    }
  }
  return primes;
}
long long getpow2(long long a, string& b, long long m) {
  long long mod = 1000000007;
  long long num = 1;
  for (long long i = b.size() - 1; i >= 0; i--) {
    if (b[i] == 1) {
      num = (num * a) % m;
    } else if (b[i] == 2) {
      num = ((num * a) % m * a) % m;
    }
    a = ((a * a) % m * a) % m;
  }
  return num % m;
}
bool check(long long i, long long m) {
  long long ret = 0;
  i = i / 5;
  while (i) {
    ret += i;
    i = i / 5;
  }
  return (ret == m);
}
long long getcode(string input, long long size, long long arr[]) {
  if (size == 0 || size == 1) {
    return 1;
  }
  if (arr[size] > 0) {
    return arr[size];
  } else {
    long long output = getcode(input, size - 1, arr);
    if ((input[size - 1] - '0') + (input[size - 2] - '0') * 10 <= 26) {
      output += getcode(input, size - 2, arr);
    }
    arr[size] = output;
    return arr[size];
  }
}
long long output[5005];
unsigned long long getcode1(string input, long long size) {
  memset(output, 0, sizeof(output));
  output[0] = 1;
  unsigned long long ans = 0;
  for (long long i = 1; i < size; i++) {
    long long tmp1 = 0;
    long long tmp2 = 0;
    if (input[i] - '0' != 0) {
      tmp1 = 1;
      output[i] = output[i - 1];
    }
    if (((input[i - 1] - '0') * 10 + (input[i] - '0') <= 26) &&
        ((input[i - 1] - '0') * 10 + (input[i] - '0') >= 10)) {
      tmp2 = 1;
      output[i] = (output[i - 2] + output[i]) % 1000000007;
    }
    if (tmp1 == 0 && tmp2 == 0) {
      output[i] = 0;
      ans = 0;
      return ans;
    }
  }
  ans = output[size - 1];
  return ans;
}
long long getans(long long* input, long long n) {
  long long* dp = new long long[n];
  dp[0] = input[0];
  dp[1] = max(input[0], input[1]);
  for (long long i = 2; i < n; i++) {
    dp[i] = max(dp[i - 2] + input[i], dp[i - 1]);
  }
  long long ans = dp[n - 1];
  delete[] dp;
  return ans;
}
long long solve(long long n) {
  if (n == 0) {
    return 0;
  }
  if (n == 1) {
    return 1;
  }
  long long ans1 = max(n / 2, solve(n / 2));
  long long ans2 = max(n / 3, solve(n / 3));
  long long ans3 = max(n / 4, solve(n / 4));
  long long ans = max(n, ans1 + ans2 + ans3);
  return ans;
}
long long mincost(long long input[][501], long long R, long long C) {
  input[R][C] = 1;
  for (long long i = R - 1; i >= 1; i--) {
    input[i][C] = input[i + 1][C] - input[i][C];
    if (input[i][C] <= 0) {
      input[i][C] = 1;
    }
  }
  for (long long i = C - 1; i >= 1; i--) {
    input[R][i] = input[R][i + 1] - input[R][i];
    if (input[R][i] <= 0) {
      input[R][i] = 1;
    }
  }
  for (long long i = R - 1; i >= 1; i--) {
    for (long long j = C - 1; j >= 1; j--) {
      input[i][j] = min(input[i + 1][j], input[i][j + 1]) - input[i][j];
      if (input[i][j] <= 0) {
        input[i][j] = 1;
      }
    }
  }
  return input[1][1];
}
float inputx[105];
float inputy[105];
float f[105];
float dp[105];
bool dp1[105];
map<string, long long> m1;
long long arrx[100001];
string s1[100001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k, m;
  cin >> n >> k >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> s1[i];
  }
  for (long long i = 1; i <= n; i++) {
    cin >> arrx[i];
    m1.insert(pair<string, long long>(s1[i], i));
  }
  for (long long i = 0; i < k; i++) {
    long long num1, num2;
    cin >> num1;
    vector<long long> vnum;
    long long ans = INT_MAX;
    for (long long j = 0; j < num1; j++) {
      cin >> num2;
      vnum.push_back(num2);
      ans = min(arrx[num2], ans);
    }
    for (long long j = 0; j < num1; j++) {
      arrx[vnum[j]] = ans;
    }
  }
  long long cost = 0;
  for (long long i = 0; i < m; i++) {
    string s2;
    cin >> s2;
    long long num = m1[s2];
    cost += arrx[num];
  }
  cout << cost << '\n';
  return 0;
}