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 maxN = 1e5 + 9, MOD = 1e9 + 7, LG = 11, BASE = 29;
const double EPS = 0;
const int sz = 1 << 17;
int n, ct, L;
vector<int> adj[maxN];
long long S, ans, w[maxN], T[1 << 18];
void updat(int w, long long x) {
  w += sz;
  T[w] = x;
  while (w > 1) w >>= 1, T[w] = T[w << 1] + T[w << 1 | 1];
}
int getind(int w, long long x) {
  if (w >= sz) {
    if (T[w] > x) return w + 1;
    return w;
  }
  if (T[w << 1 | 1] >= x) return getind(w << 1 | 1, x);
  return getind(w << 1, x - T[w << 1 | 1]);
}
int dfs(int nd, int an) {
  updat(ct++, w[nd]);
  int sta = max(ct - L, getind(1, S) - sz), mini = ct;
  for (int sn : adj[nd])
    if (sn != an) mini = min(mini, dfs(sn, nd));
  updat(ct - 1, 0);
  if (mini >= ct--) {
    ans++;
    return sta;
  }
  return mini;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> L >> S;
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    if (w[i] > S) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 1, pa; i < n; i++) {
    cin >> pa;
    pa--;
    adj[pa].push_back(i);
  }
  dfs(0, -1);
  cout << ans << '\n';
}
