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;
int w[100001];
int p[100001];
int d[100001];
bool vis[100001];
vector<int> at[100001];
int ret = 0;
void solve(int cur) {
  if (vis[cur]) return;
  ret++;
  long long weight = S;
  int cnt = L;
  while (weight - w[cur] >= 0 && cnt - 1 >= 0 && cur) {
    weight -= w[cur];
    cnt--;
    vis[cur] = 1;
    cur = p[cur];
  }
}
int main() {
  cout << setprecision(10);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> L >> S;
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    if (w[i] > S) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
    d[i] = d[p[i]] + 1;
    at[d[i]].push_back(i);
  }
  at[0].push_back(1);
  for (int dep = n; dep >= 0; dep--)
    for (auto e : at[dep]) solve(e);
  cout << ret << '\n';
  return 0;
}
