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;
template <class T>
void show(T a) {
  for (auto x : a) cout << x << " ";
  cout << endl;
}
const string DIGITS = "0123456789";
const long long N = 1e5 + 9;
long long w[N];
vector<long long> g[N];
long long ans = 0;
long long s;
long long l;
long long deep[N];
const long long LOG = 20;
long long pr[LOG][N];
long long dp[N];
void pre_dfs(long long v) {
  dp[v] = w[v];
  if (v != 1) {
    dp[v] += dp[pr[0][v]];
    deep[v] = deep[pr[0][v]] + 1;
  }
  for (long long i = 1; i < LOG; ++i) pr[i][v] = pr[i - 1][pr[i - 1][v]];
  for (long long to : g[v]) pre_dfs(to);
}
long long up(long long u, long long dist) {
  for (long long i = LOG - 1; i >= 0; --i) {
    while (dist >= (1 << i)) {
      u = pr[i][u];
      dist -= (1 << i);
    }
  }
  return u;
}
inline long long get_w(long long v, long long dist) {
  long long p = up(v, dist);
  if (p == 1) return dp[v];
  return dp[v] - dp[pr[0][p]];
}
inline long long find_h(long long v, long long weight) {
  long long l = 0, r = deep[v] + 1;
  while (r - l > 1) {
    long long m = l + r >> 1;
    if (get_w(v, m) + weight <= s)
      l = m;
    else
      r = m;
  }
  return l;
}
pair<long long, long long> dfs(long long v) {
  if (g[v].empty()) {
    ans++;
    return {1, w[v]};
  }
  long long height = -1;
  pair<long long, long long> ret;
  for (long long to : g[v]) {
    pair<long long, long long> cur = dfs(to);
    cur.second += w[v];
    ++cur.first;
    if (cur.first <= l && cur.second <= s) {
      long long cur_h = find_h(v, cur.second - w[v]);
      cur_h = min(cur_h, (l - (long long)cur.first));
      if (cur_h > height) {
        height = cur_h;
        ret = cur;
      }
    }
  }
  if (height == -1) {
    ans++;
    return {1, w[v]};
  }
  return ret;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n >> l >> s;
  for (long long i = 1; i <= n; ++i) cin >> w[i];
  for (long long v = 2; v <= n; ++v) {
    long long par;
    cin >> par;
    pr[0][v] = par;
    g[par].push_back(v);
  }
  pr[0][1] = 1;
  if (s < *max_element(w + 1, w + n + 1)) {
    cout << "-1";
    return 0;
  }
  pre_dfs(1);
  dfs(1);
  cout << ans;
}
