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;
void debug_out() { cerr << endl; }
template <class T>
ostream& prnt(ostream& out, T v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  return prnt(out, v);
}
template <class T>
ostream& operator<<(ostream& out, set<T> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ' ' << p.second << ')';
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const long long N = 100100;
long long n, lg, w[N], s, dp[N], a, ans;
vector<long long> v[N], first;
long long cb() {
  long long l = 0, r = min((long long)first.size() - 2, lg - 1);
  while (l != r) {
    long long mid = (l + r + 1) / 2;
    if (first.back() - first[first.size() - mid - 2] <= s)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
void dfs(long long node, long long tot) {
  first.push_back(tot);
  long long up = -1;
  for (auto i : v[node]) {
    dfs(i, tot + w[i]);
    up = max(up, dp[i] - 1);
  }
  if (up == -1)
    dp[node] = cb(), ans++;
  else
    dp[node] = up;
  first.pop_back();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> lg >> s;
  for (long long i = 1; i <= n; i++) {
    cin >> w[i];
    if (w[i] > s) return cout << "-1\n", 0;
  }
  for (long long i = 2; i <= n; i++) {
    cin >> a;
    v[a].push_back(i);
  }
  first.push_back(0);
  dfs(1, w[1]);
  cout << ans << '\n';
}
