Dlsj is competing in a contest with n (0 < n \le 20)n(0<n≤20) problems. And he knows the answer of all of these problems.

However, he can submit ii-th problem if and only if he has submitted (and passed, of course) s_isi​ problems, the p_{i, 1}pi,1​-th, p_{i, 2}pi,2​-th, ......, p_{i, s_i}pi,si​​-th problem before.(0 < p_{i, j} \le n,0 < j \le s_i,0 < i \le n)(0<pi,j​≤n,0<j≤si​,0<i≤n) After the submit of a problem, he has to wait for one minute, or cooling down time to submit another problem. As soon as the cooling down phase ended, he will submit his solution (and get "Accepted" of course) for the next problem he selected to solve or he will say that the contest is too easy and leave the arena.

"I wonder if I can leave the contest arena when the problems are too easy for me." "No problem." —— CCF NOI Problem set

If he submits and passes the ii-th problem on tt-th minute(or the tt-th problem he solve is problem ii), he can get t \times a_i + b_it×ai​+bi​ points. (|a_i|, |b_i| \le 10^9)(∣ai​∣,∣bi​∣≤109).

Your task is to calculate the maximum number of points he can get in the contest.

Input

The first line of input contains an integer, nn, which is the number of problems.

Then follows nn lines, the ii-th line contains s_i + 3si​+3 integers, a_i,b_i,s_i,p_1,p_2,...,p_{s_i}ai​,bi​,si​,p1​,p2​,...,psi​​as described in the description above.

Output

Output one line with one integer, the maximum number of points he can get in the contest.

Hint

In the first sample.

On the first minute, Dlsj submitted the first problem, and get 1 \times 5 + 6 = 111×5+6=11 points.

On the second minute, Dlsj submitted the second problem, and get 2 \times 4 + 5 = 132×4+5=13 points.

On the third minute, Dlsj submitted the third problem, and get 3 \times 3 + 4 = 133×3+4=13 points.

On the forth minute, Dlsj submitted the forth problem, and get 4 \times 2 + 3 = 114×2+3=11 points.

On the fifth minute, Dlsj submitted the fifth problem, and get 5 \times 1 + 2 = 75×1+2=7 points.

So he can get 11+13+13+11+7=5511+13+13+11+7=55 points in total.

In the second sample, you should note that he doesn't have to solve all the problems.

樣例輸入1複製

5
5 6 0
4 5 1 1
3 4 1 2
2 3 1 3
1 2 1 4

樣例輸出1複製

55

樣例輸入2複製

1
-100 0 0

樣例輸出2複製

0

題目來源

題意：A一道題有前提條件是A掉另外一些給定的題，每一道題有屬性a和b，在t時刻A掉一道題的得分是a*t+b，求最大得分。

思路：題數小於20，考慮狀壓dp，AC一道題的條件就是狀態轉移的條件。用二進位制數來表示答題情況，1代表已經AC，那麼首先對於每一道題，用二進位制數來代表要A掉它的條件，然後dp[i]代表i這個條件的最大得分，那麼轉移方程就是dp[i]=max(dp[i^(1<<k)]+a*num+b)(0<k<n)，條件是狀態i的從右往左第k位0也就是第k題還沒有AC，同時狀態i滿足第k題的前置條件。

#include<bits/stdc++.h>
using namespace std;
const int N = 20 + 5;
typedef long long ll;
struct A {
    ll a, b;
    int s;
} p[N];
ll dp[(1<<20) + 5];
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; ++ i) {
        scanf("%lld%lld", &p[i].a, &p[i].b);
        p[i].s = 0; int k;
        scanf("%d", &k);
        for(int j = 0; j < k; ++ j) {
            int pre; scanf("%d", &pre);
            p[i].s |= (1 << pre-1);
        }
    }
    ll ans = 0;
    memset(dp, -1, sizeof(dp));
    dp[0] = 0;
    for(int s = 1; s < (1 << n); ++ s) {
        int num = 0;
        for(int i = 0; i < n; ++ i) if(s & (1 << i))
            num ++;
        for(int i = 0; i < n; ++ i) {
            if((s & (1 << i)) && (s & (p[i].s)) == p[i].s && dp[s ^ (1 << i)] != -1) {
                dp[s] = max(dp[s], dp[s ^ (1 << i)] + p[i].a * num + p[i].b);
            }
        }
        ans = max(ans, dp[s]);
    }
    printf("%lld\n", ans);
}