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 maxn = 100010;
int n, L, cnt;
long long S, w[maxn], ans[maxn];
vector<int> G[maxn];
vector<long long> cur;
void dfs(int v) {
  long long mx = 0;
  for (int u : G[v]) {
    cur.push_back(cur.back() + w[u]), dfs(u), cur.pop_back();
    mx = max(mx, ans[u]);
  }
  if (mx >= w[v]) {
    ans[v] = mx - w[v];
  } else {
    cnt++, ans[v] = min(S, cur.back() - cur[cur.size() - L - 1]) - w[v];
  }
}
int main() {
  scanf("%d %d %lld", &n, &L, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    if (w[i] > S) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 2, p; i <= n; i++) {
    scanf("%d", &p), G[p].push_back(i);
  }
  cur = vector<long long>(L, 0), cur.push_back(w[1]), dfs(1);
  printf("%d\n", cnt);
  return 0;
}
