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;
constexpr int ls = 18;
constexpr int ms = 100010;
vector<int> g[ms];
vector<int> h(ms, -1);
int n, L, x;
long long S, V[ms], valorAcumulado[ms];
int anc[ls][ms], res;
pair<long long, int> dp[ms];
void dfs(int root) {
  valorAcumulado[root] += V[root];
  for (const int& x : g[root]) {
    if (h[x] == -1) {
      valorAcumulado[x] += valorAcumulado[root];
      h[x] = h[root] + 1;
      dfs(x);
    }
  }
}
void preCalc() {
  anc[0][0] = -1;
  for (int lvl = 1; lvl < ls; ++lvl) {
    for (int node = 0; node < n; ++node) {
      if (h[node] < (1 << lvl))
        anc[lvl][node] = -1;
      else
        anc[lvl][node] = anc[lvl - 1][anc[lvl - 1][node]];
    }
  }
}
int calculaPulo(int node) {
  long long sobra = S - dp[node].first;
  int cur = node;
  for (int pot = 17; pot >= 0; --pot) {
    if (anc[pot][cur] != -1) {
      int cand = anc[pot][cur];
      if (h[node] - h[cand] + dp[node].second > L) continue;
      long long incremento = valorAcumulado[cur] - V[cur];
      if (anc[0][cand] != -1) incremento -= valorAcumulado[anc[0][cand]];
      if (incremento <= sobra) {
        sobra -= incremento;
        cur = cand;
      }
    }
  }
  return h[cur];
}
void solve(int node) {
  int best = h[node] + 1;
  long long acumula = V[node];
  for (const int kid : g[node]) {
    if (h[kid] > h[node]) {
      solve(kid);
      int up = calculaPulo(kid);
      if (up < best) {
        best = up;
        acumula = (V[node] + dp[kid].first);
        dp[node].second = (1 + dp[kid].second);
      }
    }
  }
  if (acumula == V[node]) ++res;
  dp[node].first = acumula;
  dp[node].second = max(dp[node].second, 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> L >> S;
  for (int i = 0; i < n; ++i) {
    cin >> V[i];
    if (V[i] > S) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 1; i < n; ++i) {
    cin >> x;
    --x;
    g[i].push_back(x);
    g[x].push_back(i);
    anc[0][i] = x;
  }
  h[0] = 0;
  dfs(0);
  preCalc();
  solve(0);
  cout << res << endl;
  return 0;
}
