You are given a rooted tree on n vertices, its root is the vertex number 1. The i-th vertex contains a number w_i. Split it into the minimum possible number of vertical paths in such a way that each path contains no more than L vertices and the sum of integers w_i on each path does not exceed S. Each vertex should belong to exactly one path.

A vertical path is a sequence of vertices v_1, v_2, …, v_k where v_i (i ≥ 2) is the parent of v_{i - 1}.

Input

The first line contains three integers n, L, S (1 ≤ n ≤ 10^5, 1 ≤ L ≤ 10^5, 1 ≤ S ≤ 10^{18}) — the number of vertices, the maximum number of vertices in one path and the maximum sum in one path.

The second line contains n integers w_1, w_2, …, w_n (1 ≤ w_i ≤ 10^9) — the numbers in the vertices of the tree.

The third line contains n - 1 integers p_2, …, p_n (1 ≤ p_i < i), where p_i is the parent of the i-th vertex in the tree.

Output

Output one number — the minimum number of vertical paths. If it is impossible to split the tree, output -1.

Examples

Input

3 1 3
1 2 3
1 1


Output

3

Input

3 3 6
1 2 3
1 1


Output

2

Input

1 1 10000
10001


Output

-1

Note

In the first sample the tree is split into \{1\},\ \{2\},\ \{3\}.

In the second sample the tree is split into \{1,\ 2\},\ \{3\} or \{1,\ 3\},\ \{2\}.

In the third sample it is impossible to split the tree.
#include <bits/stdc++.h>
using namespace std;
set<pair<long long, int> > pos[200309];
vector<int> g[200309];
int s[200309];
long long S, w[200309], off[200309];
int n, L;
void merge(int u, int v) {
  if (pos[v].size() > pos[u].size()) {
    swap(pos[u], pos[v]);
    swap(off[u], off[v]);
  }
  for (auto pp : pos[v]) {
    long long curw = pp.first + off[v] - off[u];
    pos[u].insert({curw, pp.second});
  }
  pos[v].clear();
}
void dfs(int u, int h) {
  s[u] = 0;
  off[u] = 0;
  for (int v : g[u]) {
    dfs(v, h + 1);
    s[u] += s[v];
    off[v] += w[u];
    merge(u, v);
  }
  while (!pos[u].empty()) {
    auto cur = *pos[u].rbegin();
    if (cur.first + off[u] > S)
      pos[u].erase(cur);
    else
      break;
  }
  while (!pos[u].empty()) {
    auto cur = *pos[u].rbegin();
    if (cur.second - h > L - 1)
      pos[u].erase(cur);
    else
      break;
  }
  if (pos[u].empty()) {
    s[u]++;
    pos[u].insert({w[u], h});
    off[u] = 0;
  }
}
int main() {
  scanf("%d %d %lld", &n, &L, &S);
  bool ok = true;
  for (int u = 1; (u) <= int(n); (u)++) {
    scanf("%lld", &w[u]);
    if (w[u] > S) ok = false;
  }
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    g[p].push_back(i);
  }
  dfs(1, 0);
  printf("%d\n", ok ? s[1] : -1);
  return 0;
}
