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, ok = 1, sus[100005], a[100005][30], niv[100005], t[100005], nrlanturi;
long long cost[100005], s, sum[100005], inf = 1e18;
vector<int> v[100005];
void dfs1(int nod) {
  sum[nod] = sum[t[nod]] + cost[nod];
  niv[nod] = niv[t[nod]] + 1;
  sus[nod] = nod;
  a[nod][0] = t[nod];
  for (auto it = v[nod].begin(); it != v[nod].end(); ++it) dfs1(*it);
}
void blf() {
  int i, j;
  for (i = 1; i <= 20; i++)
    for (j = 1; j <= n; j++) a[j][i] = a[a[j][i - 1]][i - 1];
}
void dfs2(int nod) {
  int lung = l - 1;
  for (int i = 20; i >= 0; i--)
    if ((lung >= (1 << i)) &&
        (sum[nod] - sum[a[sus[nod]][i]] + cost[a[sus[nod]][i]] <= s)) {
      lung -= (1 << i);
      sus[nod] = a[sus[nod]][i];
    }
  for (auto it = v[nod].begin(); it != v[nod].end(); ++it) dfs2(*it);
}
void citire() {
  int i;
  cin >> n >> l >> s;
  for (i = 1; i <= n; i++) {
    cin >> cost[i];
    if (cost[i] > s) ok = 0;
  }
  for (i = 2; i <= n; i++) {
    cin >> t[i];
    v[t[i]].push_back(i);
  }
}
long long DFS(int nod) {
  long long ans = inf;
  for (auto it = v[nod].begin(); it != v[nod].end(); ++it)
    ans = min(DFS(*it), ans);
  if (ans > niv[nod]) {
    nrlanturi++;
    return niv[sus[nod]];
  } else
    return ans;
}
int main() {
  citire();
  if (!ok) {
    cout << -1;
    return 0;
  }
  dfs1(1);
  blf();
  dfs2(1);
  DFS(1);
  cout << nrlanturi;
  return 0;
}
