// 树上背包(后序遍历优化)，时间复杂度O(nm)

// https://loj.ac/p/160

#include <bits/stdc++.h>

using namespace std;

const int N = 50003;

const int M = 60003;

int n, m, tot;

int w[N], v[N], s[N], dfn[N], rdfn[N];

vector<vector<int>> g, f;

void dfs(int u) {

s[u] = 1;

for (int v : g[u]) {

dfs(v);

s[u] += s[v];

}

dfn[u] = ++tot; // 求后序遍历序

rdfn[tot] = u;

}

int main() {

ios::sync_with_stdio(false);

cin.tie(nullptr);

cin >> n >> m;

g.resize(n + 3);

f.resize(n + 3, vector<int>(m + 1));

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

int d;

cin >> d;

g[d].push_back(i);

}

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

cin >> w[i];

}

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

cin >> v[i];

}

dfs(0);

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

for (int j = 0; j <= m; ++j) {

f[i][j] = f[i - s[rdfn[i]]][j]; // 不取当前节点，则其子树不能取

if (j >= w[rdfn[i]]) { // 取当前节点

f[i][j] = max(f[i][j], f[i - 1][j - w[rdfn[i]]] + v[rdfn[i]]);

}

}

}

cout << f[n + 1][m] << "\n";

return 0;

}