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;
long long n, l, sm, ch = 0, w[101000], p[1010000][21];
vector<long long> v[101000];
long long dfs(long long s) {
  w[s] += w[p[s][0]];
  for (long long i = 1; i <= 20; i++)
    if (p[s][i - 1] != -1) p[s][i] = p[p[s][i - 1]][i - 1];
  long long res = 0;
  for (long long i = 0; i < v[s].size(); i++) res = max(res, dfs(v[s][i]));
  if (res) return res - 1;
  ch++;
  long long u = s, len = 0;
  for (long long i = 20; i >= 0; i--)
    if (p[u][i] != -1 && (w[s] - w[p[u][i]]) <= sm) {
      u = p[u][i];
      len += 1LL << i;
    }
  return min(len, l) - 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long x, f = 0;
  cin >> n >> l >> sm;
  memset(p, -1, sizeof(p));
  for (long long i = 1; i <= n; i++) {
    cin >> w[i];
    if (w[i] > sm) f = 1;
  }
  p[1][0] = 0;
  for (long long i = 1; i <= n - 1; i++) {
    cin >> x;
    v[x].push_back(i + 1);
    p[i + 1][0] = x;
  }
  dfs(1);
  if (!f)
    cout << ch << endl;
  else
    cout << -1 << endl;
}
