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;
const int N = 1e5 + 10;
vector<long long> G[N];
long long n, L, S;
long long w[N];
vector<pair<long long, long long> > f[N];
int main() {
  cin >> n >> L >> S;
  for (long long i = 1; i <= n; i++) {
    cin >> w[i];
    if (w[i] > S) return puts("-1"), 0;
  }
  for (long long i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    G[p].push_back(i);
  }
  long long ans = 0;
  for (int i = n; i; --i) {
    for (auto j : G[i])
      for (auto p : f[j])
        if (p.first + 1 <= L && p.second + w[i] <= S)
          f[i].push_back({p.first + 1, p.second + w[i]});
    if (f[i].empty()) f[i].push_back({1, w[i]}), ++ans;
  }
  cout << ans << endl;
}
