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 mod = 1e9 + 7;
const int mod2 = 1e9 + 9;
inline int pow_(long long a, int n, int p = mod) {
  int r = 1;
  while (n) {
    if (n & 1) r = r * a % p;
    n >>= 1;
    a = a * a % p;
  }
  return r;
}
int main(void) {
  int n, l;
  long long s;
  cin >> n >> l >> s;
  int w[n + 1], p[n + 1];
  memset(p, 0, sizeof(p));
  memset(w, 0, sizeof(w));
  for (int i = 1; i <= n; i++) cin >> w[i];
  for (int i = 2; i <= n; i++) cin >> p[i];
  bool done[n + 1];
  memset(done, false, sizeof(done));
  for (int i = 1; i <= n; i++)
    if (w[i] > s) return 0 * printf("-1\n");
  int ans = 0;
  for (int i = n; i >= 1; i--) {
    if (done[i]) continue;
    ans++;
    int cur = i;
    int L = l;
    long long S = s;
    while (cur and L > 0 and S >= w[cur]) {
      done[cur] = true;
      S -= w[cur];
      L--;
      cur = p[cur];
    }
  }
  cout << ans << "\n";
}
