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 MOD = 1e9 + 7;
long long n, l, s;
long long w[100000 + 10];
long long p[100000 + 10];
vector<long long> child[100000 + 10];
long long dp1[100000 + 10];
set<pair<long long, long long> > dp2[100000 + 10];
void dfs(long long x) {
  long long sum = 0;
  set<pair<long long, long long> > temp;
  for (auto it = child[x].begin(); it != child[x].end(); it++) {
    dfs((*it));
    sum += dp1[(*it)];
    for (auto it1 = dp2[(*it)].begin(); it1 != dp2[(*it)].end(); it1++) {
      temp.insert((*it1));
    }
  }
  bool valid = 0;
  for (auto it = temp.begin(); it != temp.end(); it++) {
    if ((*it).first + w[x] <= s and (*it).second < l) {
      valid = 1;
      break;
    }
  }
  if (!valid) {
    dp2[x].insert(make_pair(w[x], 1));
    dp1[x] = sum + 1;
    return;
  }
  dp1[x] = sum;
  long long len = n + 1;
  for (auto it = temp.begin(); it != temp.end(); it++) {
    if ((*it).second < min(len, l) and (*it).first + w[x] <= s) {
      len = (*it).second;
      dp2[x].insert(make_pair((*it).first + w[x], (*it).second + 1));
    }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> l >> s;
  bool check = 0;
  for (long long i = 1; i < n + 1; i++) {
    cin >> w[i];
    if (w[i] > s) {
      check = 1;
    }
  }
  for (long long i = 2; i < n + 1; i++) {
    cin >> p[i];
    child[p[i]].push_back(i);
  }
  if (check) {
    cout << -1 << '\n';
    return 0;
  }
  dfs(1);
  cout << dp1[1] << '\n';
  return 0;
}
