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 iinf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5;
int n, len, ans;
long long sum, w[N];
vector<int> e[N];
void adde(int u, int v) { e[u].push_back(v), e[v].push_back(u); }
int fa[20][N];
long long sm[20][N];
int Dfs(int u = 0, int fa = -1) {
  int mx = 0;
  for (int v : e[u])
    if (v != fa) mx = max(mx, Dfs(v, u) - 1);
  if (!mx) {
    ans++;
    long long ts = 0;
    for (int i = (20) - 1, I = (0) - 1; i > I; i--)
      if (ts + sm[i][u] <= sum) ts += sm[i][u], mx += (1 << i);
    mx = min(mx, len);
    if (!mx) ans = -iinf;
  }
  return mx;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> len >> sum;
  for (int i = (0), I = (n); i < I; i++) cin >> w[i], sm[0][i] = w[i];
  for (int i = (0), I = (20); i < I; i++)
    for (int u = (0), I = (n); u < I; u++) fa[i][u] = -1;
  for (int i = (1), I = (n); i < I; i++) {
    cin >> fa[0][i], --fa[0][i];
    adde(fa[0][i], i);
  }
  for (int i = (0), I = (19); i < I; i++)
    for (int u = (0), I = (n); u < I; u++)
      if (~fa[i][u]) fa[i + 1][u] = fa[i][fa[i][u]];
  for (int i = (0), I = (19); i < I; i++)
    for (int u = (0), I = (n); u < I; u++) {
      sm[i + 1][u] = sm[i][u];
      if (~fa[i][u]) sm[i + 1][u] += sm[i][fa[i][u]];
    }
  Dfs();
  if (ans < 0)
    cout << "-1" << '\n';
  else
    cout << ans << '\n';
  return 0;
}
