#include <bits/stdc++.h>
using namespace std;


int bit[22];

int a[22];
int b[22];
int state[22];

long long dp[1<<20];
int numbit[1<<20];
const long long INF = 1000000000000000LL;

int main() {
  bit[0] = 1;
  for (int i = 1; i < 22; i++)
    bit[i] = bit[i-1]<<1;   //bit[i] = pow(2,i);  
  numbit[0] = 0;
  for (int i = 1; i < bit[20]; i++) {
    numbit[i] = 1 + numbit[i&(i-1)];     
  }
  int n;
  while(scanf("%d", &n) == 1) {
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &a[i], &b[i]);
      int s;
      scanf("%d", &s);
      int tmp = 0;
      state[i] = 0;
      while (s--) {
        scanf("%d", &tmp);
        state[i] |= bit[tmp-1];   //2進位制位的1表示第1題需要的前題(位置==題號)  
      }
    }
      dp[0] = 0;
      for (int i = 1; i < bit[n]; i++)dp[i] = -INF;
      long long ans = 0;
      for (int i = 0; i < bit[n]; i++) {
        if (dp[i] == -INF)continue;
        ans = max(ans, dp[i]);
        for (int j = 0; j < n; j++) {
          if (i & bit[j])continue;    //做過的題跳過
          if ((i&state[j]) != state[j])continue;//  對於沒做過的題，若滿足其所有前題，則可以嘗試做或不做 注意所有位運算操作都比一般運算優先順序低，若要先進行位運算必須加()！！！
          dp[i|bit[j]] = max(dp[i|bit[j]],dp[i] + (long long)(numbit[i]+1)*a[j] + b[j]);  //做bit[j]二進位制1位置的題，保留最佳狀態
        }
      }
      cout<<ans<<endl;

  }
  return 0;
}