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 long long N = 1e5 + 7;
long long w[N], n, l, s, sum[N], cnt[N];
vector<long long> adj[N], z;
long long dfs(long long u) {
  z.push_back(z.back() + w[u]);
  vector<pair<long long, long long> > t;
  long long ans = 0;
  for (long long v : adj[u]) {
    ans += dfs(v);
    if (cnt[v] == l || sum[v] + w[u] > s) continue;
    long long k = z.size(), L = 1, R = min(l - cnt[v], k - 1);
    while (L < R) {
      long long mid = (L + R + 1) / 2;
      if (z.back() - z[k - mid - 1] <= s)
        L = mid;
      else
        R = mid - 1;
    }
    t.push_back(make_pair(L, v));
  }
  z.pop_back();
  if (t.empty()) {
    sum[u] = w[u];
    cnt[u] = 1;
    return ans + adj[u].size();
  } else {
    auto [x, v] = *max_element(t.begin(), t.end());
    sum[u] = sum[v] + w[u];
    cnt[u] = cnt[v] + 1;
    return ans + adj[u].size() - 1;
  }
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  z.push_back(0);
  cin >> n >> l >> s;
  for (long long i = 1; i <= n; i++) cin >> w[i];
  if (*max_element(w + 1, w + n + 1) > s) {
    cout << -1 << endl;
    return 0;
  }
  for (long long i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    adj[p].push_back(i);
  }
  cout << 1 + dfs(1) << endl;
}
