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;
int n, l;
long long s;
vector<int> edges[100005];
int par[100005][17];
long long st[100005][17];
int ans = 0;
bool fuck = 0;
int dfs(int u) {
  int mx = 0;
  for (auto v : edges[u]) {
    mx = max(mx, dfs(v));
  }
  if (!mx) {
    long long ws = 0;
    int t = u, cnt = 0;
    for (int i = (16); i >= (0); i--) {
      if (ws + st[t][i] > s) continue;
      ws += st[t][i];
      t = par[t][i];
      cnt += (1 << i);
    }
    cnt = min(cnt, l);
    if (!cnt)
      fuck = 1;
    else
      mx = cnt, ans++;
  }
  return (mx - 1);
}
int main() {
  scanf("%d %d %lld", &n, &l, &s);
  for (int i = (1); i <= (n); i++) {
    scanf("%lld", &st[i][0]);
  }
  for (int i = (2); i <= (n); i++) {
    int p;
    scanf("%d", &p);
    edges[p].push_back(i);
    par[i][0] = p;
  }
  for (int i = (1); i <= (16); i++) {
    for (int j = (1); j <= (n); j++) {
      par[j][i] = par[par[j][i - 1]][i - 1];
      st[j][i] = st[j][i - 1] + st[par[j][i - 1]][i - 1];
    }
  }
  dfs(1);
  printf("%d\n", fuck ? -1 : ans);
  return 0;
}
