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;
using LL = long long;
using PII = pair<int, int>;
LL inv(LL x, LL m) { return x > 1 ? inv(m % x, m) * (m - m / x) % m : x; }
LL powmod(LL a, LL k, LL m) {
  LL res = 1;
  for (a %= m; k; k >>= 1, a = a * a % m)
    if (k & 1) res = res * a % m;
  return res;
}
int main() {
  int n, L;
  LL S;
  scanf("%d%d%lld", &n, &L, &S);
  vector<LL> W(n + 1);
  for (int i = (1); i <= (n); ++i) scanf("%lld", &W[i]);
  for (int i = (1); i <= (n); ++i)
    if (W[i] > S) return puts("-1"), 0;
  vector<vector<int>> G(n + 1), Fa(n + 1, vector<int>(18));
  vector<int> Dep(n + 1);
  vector<LL> Sum(n + 1);
  for (int i = (2); i <= (n); ++i) {
    int p;
    scanf("%d", &p);
    G[p].push_back(i);
    G[i].push_back(p);
  }
  function<void(int, int, int, LL)> dfs = [&](int u, int p, int dep, LL sum) {
    Fa[u][0] = p;
    Dep[u] = dep;
    Sum[u] = sum + W[u];
    for (int i = 1; i < 18; i++) Fa[u][i] = Fa[Fa[u][i - 1]][i - 1];
    for (auto v : G[u])
      if (v != p) {
        dfs(v, u, dep + 1, Sum[u]);
      }
  };
  dfs(1, 0, 1, 0);
  vector<int> F(n + 1);
  vector<LL> H(n + 1);
  int ans = 0;
  function<void(int, int)> dfs2 = [&](int u, int p) {
    F[u] = 1;
    H[u] = W[u];
    int ma = 0;
    for (auto v : G[u])
      if (v != p) {
        dfs2(v, u);
        LL rsum = S - H[v];
        int rcnt = L - F[v];
        int tp = u;
        for (int i = 17; i >= 0; i--) {
          int tmp = Fa[tp][i];
          if (Sum[u] - Sum[tmp] <= rsum) tp = tmp;
        }
        int can = min(rcnt, Dep[u] - Dep[tp]);
        if (can > ma) {
          F[u] = F[v] + 1;
          H[u] = H[v] + W[u];
          ma = can;
        }
      }
    if (ma == 0) ans++;
  };
  dfs2(1, -1);
  printf("%d\n", ans);
  return 0;
}
