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 pres[100005], p[100005], w[100005], s, l, n, h[100005], res;
int vis[100005];
vector<int> ed[100005];
void dfs(int i, int p) {
  for (int j : ed[i]) {
    if (j == p) continue;
    h[j] = h[i] + 1;
    pres[j] = pres[i] + w[j];
    dfs(j, i);
  }
}
void solve() {
  cin >> n >> l >> s;
  for (int i = 0; i < n; i++) cin >> w[i];
  for (int i = 0; i < n; i++)
    if (w[i] > s) {
      cout << -1;
      return;
    }
  for (int i = 0; i < n - 1; i++) {
    int pa;
    cin >> pa;
    p[i + 1] = pa - 1;
    ed[pa - 1].push_back(i + 1);
    ed[i + 1].push_back(pa - 1);
  }
  p[0] = -1;
  dfs(0, -1);
  pair<int, int> order[n];
  for (int i = 0; i < n; i++) order[i] = pair<int, int>(h[i], i);
  sort(order, order + n, greater<pair<int, int> >());
  for (pair<int, int> pp : order) {
    int v = pp.second;
    if (vis[v]) continue;
    res++;
    vis[v] = 1;
    int end = v, p1 = v;
    long long wt, ht;
    while (end != -1) {
      wt = pres[v] - pres[end] + w[end];
      ht = h[v] - h[end] + 1;
      if (wt > s || ht > l) break;
      vis[end] = 1;
      end = p[end];
    }
    while (v != -1 && p[v] != end) {
      p1 = v;
      v = p[v];
      p[p1] = end;
    }
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}
