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;
long long n, l, s;
long long wa[200000];
vector<long long> pref;
vector<int> sas[200000];
int parents[200000];
int wynik;
class pat {
 public:
  long long l, s;
};
int bin(pat p) {
  int kon = pref.size();
  int pon = 1;
  while (kon != pon) {
    int sr = (kon + pon) / 2;
    if (pref[pref.size() - 1] - pref[sr - 1] + p.s > s)
      pon = sr + 1;
    else
      kon = sr;
  }
  return pref.size() - pon;
}
pat kiczak(int w) {
  vector<pat> v;
  pref.push_back(pref[pref.size() - 1] + wa[w]);
  for (auto s : sas[w]) v.push_back(kiczak(s));
  pat best;
  int high = 0;
  best.l = -1;
  best.s = -1;
  for (int i = 0; i < v.size(); i++) {
    long long x = bin(v[i]);
    x = min(x, l - v[i].l);
    if (x > high) {
      high = x;
      best.s = v[i].s + wa[w];
      best.l = v[i].l + 1;
    }
  }
  if (high > 0) {
    return best;
  }
  wynik++;
  pat ret;
  ret.l = 1;
  ret.s = wa[w];
  pref.pop_back();
  return ret;
}
int main() {
  cin >> n >> l >> s;
  pref.push_back(0);
  for (int i = 1; i <= n; i++) {
    cin >> wa[i];
    if (wa[i] > s) {
      cout << "-1";
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    parents[x] = i;
    sas[x].push_back(i);
  }
  kiczak(1);
  cout << wynik;
  return 0;
}
