1 # include<iostream>

 2 # include<bits/stdc++.h>

 3 using namespace std;

 4 # define int long long

 5 # define endl "\n"

 6 const int N = 2e5 + 10;

 7 vector<int> g[N];

 8 int s[N];

 9 int c[N];

10 map<pair<int, int>, int> mp;

11 int f[N][2];

12 int d[N];

13 int dfs2(int u, int F) {

14     d[u] = g[u].size();

15     if(d[u] == 0) return F*s[u];

16     if (mp.count({u, F})) return mp[ {u, F}];//当前状态是否被搜索过

17     int f0 = F / d[u];//平分状态的路径数

18     int re = F % d[u];

19     int sum = F * s[u];

20     int f1 = (F + d[u] - 1) / d[u];//多获取状态的路径数

21     priority_queue<int> q;

22     for (auto v : g[u]) {

23         f[v][0] = dfs2(v, f0);

24         if (f0 == f1) f[v][1] = f[v][0];

25         else f[v][1] = dfs2(v, f1);

26         sum += f[v][0];

27         q.push(f[v][1] - f[v][0]);//优先队列保存的是多出来的路径所能产生的最大贡献

28     }

29     while (re--) {

30         sum += q.top();

31         q.pop();

32     }

33     return mp[ {u, F}] = sum;

34

35 }

36 void solve() {

37     int n, k;

38     cin >> n >> k;

39     mp.clear();

40     for (int i = 1; i <= n; ++i) {

41         g[i].clear();

42         f[i][1] = f[i][0] = 0;

43     }

44     for (int i = 2; i <= n; ++i) {

45         int x;

46         cin >> x;

47         g[x].push_back(i);

48     }

49     for (int i = 1; i <= n; ++i) cin >> s[i];

50     cout << dfs2(1, k) << endl;

51

52 }

53 int tt;

54 signed main() {

55     ios::sync_with_stdio(false);

56     cin.tie(0);

57     cout.tie(0);

58     tt = 1;

59

61     cin >> tt;

62     while (tt--)solve();

63

64

65     return 0;

66 }