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 INF = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5 + 10;
int n, L;
long long S;
int p[N];
vector<int> g[N];
long long w[N];
vector<long long> anc;
int ans;
int dfs2(int u) {
  int dep = anc.size();
  anc.push_back(anc.back() + w[u]);
  int mn = INF;
  for (auto v : g[u]) {
    mn = min(mn, dfs2(v));
  }
  if (mn < dep) {
    anc.pop_back();
    return mn;
  }
  int l = 0, r = anc.size();
  long long sum = anc.back();
  while (l < r) {
    int mid = l + (r - l) / 2;
    if (sum - anc[mid] <= S && (int)anc.size() - mid - 1 <= L) {
      r = mid;
    } else {
      l = mid + 1;
    }
  }
  ++ans;
  anc.pop_back();
  return l;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> L >> S;
  for (int i = 1; i <= n; ++i) {
    cin >> w[i];
    if (w[i] > S) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 2; i <= n; ++i) {
    cin >> p[i];
    g[p[i]].push_back(i);
  }
  anc.push_back(0);
  dfs2(1);
  cout << ans << '\n';
  return 0;
}
