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 + 7;
int w[N], p[N], sz[N];
vector<int> cld[N];
bool dead[N];
int st[N], en[N], dep[N];
long long sum[N];
int tim = 0;
void dfs(int u) {
  st[u] = ++tim;
  sum[u] = w[u] + sum[p[u]];
  dep[u] = 1 + dep[p[u]];
  sz[u] = 1;
  for (int v : cld[u]) {
    dfs(v);
    sz[u] += sz[v];
  }
  en[u] = tim;
}
int ft[N];
void add(int idx) {
  for (idx = idx; idx < N; idx += idx & -idx) ft[idx]++;
}
int getsum(int idx) {
  int ans = 0;
  for (idx = idx; idx > 0; idx -= idx & -idx) ans += ft[idx];
  return ans;
}
int rep[N];
void init() {
  for (int i = 0; i < N; i++) rep[i] = i;
}
int find(int u) {
  if (rep[u] == u) return u;
  return rep[u] = find(rep[u]);
}
void erase(int u) {
  assert(!dead[u]);
  dead[u] = 1;
  rep[find(u)] = rep[find(p[u])];
  add(st[u]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, x;
  long long z;
  cin >> n >> x >> z;
  for (int i = 1; i <= n; i++) cin >> w[i];
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
    cld[p[i]].push_back(i);
  }
  init();
  dfs(1);
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (cld[i].size() == 0) q.push(i);
  int ans = 0;
  while (q.size()) {
    int u = q.front();
    q.pop();
    ans++;
    int cur = u;
    while (cur) {
      assert(!dead[cur]);
      int aa = dep[u] - dep[p[cur]];
      long long bb = sum[u] - sum[p[cur]];
      if (aa > x || bb > z) break;
      erase(cur);
      cur = find(p[cur]);
    }
    if (cur == u) {
      cout << -1;
      return 0;
    }
    if (getsum(en[cur]) - getsum(st[cur] - 1) == sz[cur] - 1) q.push(cur);
  }
  cout << ans << endl;
}
