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 P = 1e9 + 7;
int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a < 0 ? a + P : a;
}
int mul(int a, int b) { return 1ll * a * b % P; }
int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 1e5 + 7;
int n, w[N], L, d[N], u, fa[N], f[N], vis[N], ans;
long long s[N], S;
priority_queue<pair<long long, int> > q;
vector<int> g[N];
int get(int x) { return x == f[x] ? x : f[x] = get(f[x]); }
void dfs(int u) {
  for (auto v : g[u]) {
    d[v] = d[u] + 1;
    s[v] = s[u] + w[v];
    dfs(v);
  }
  q.push(make_pair(s[u], u));
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> n >> L >> S;
  for (int i = (1); i < (n + 1); i++) {
    cin >> w[i];
    if (w[i] > S) {
      cout << -1;
      return 0;
    }
  }
  s[1] = w[1];
  for (int i = (2); i < (n + 1); i++) {
    cin >> u;
    g[u].push_back(i);
    fa[i] = u;
  }
  dfs(1);
  for (int i = (1); i < (n + 1); i++) f[i] = i;
  while (!q.empty()) {
    int x = q.top().second, y = x;
    q.pop();
    if (vis[x]) continue;
    ans++;
    while (x && d[y] - d[x] + 1 <= L && s[y] - s[x] + w[x] <= S) {
      vis[x] = 1;
      f[x] = get(fa[x]);
      x = f[x];
    }
  }
  cout << ans;
  return 0;
}
