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;
int L;
long long S;
int c[100005];
vector<int> G[100005];
int offsetLen[100005];
long long offsetVal[100005];
multiset<pair<int, long long> > s[100005];
int ans = 0;
void dfs(int u) {
  for (auto v : G[u]) {
    dfs(v);
    if (s[v].size() > s[u].size()) {
      s[u].swap(s[v]);
      swap(offsetLen[u], offsetLen[v]);
      swap(offsetVal[u], offsetVal[v]);
    }
  }
  offsetLen[u]++;
  offsetVal[u] += c[u];
  for (auto v : G[u]) {
    for (auto vv : s[v]) {
      s[u].insert({vv.first + offsetLen[v] + 1 - offsetLen[u],
                   vv.second + offsetVal[v] + c[u] - offsetVal[u]});
    }
  }
  while (s[u].size() && (*s[u].rbegin()).first + offsetLen[u] > L) {
    s[u].erase(s[u].find(*s[u].rbegin()));
    ans++;
  }
  while (s[u].size() && (*s[u].rbegin()).second + offsetVal[u] > S) {
    s[u].erase(s[u].find(*s[u].rbegin()));
    ans++;
  }
  if (s[u].empty()) {
    s[u].insert({1 - offsetLen[u], c[u] - offsetVal[u]});
  }
}
int main() {
  scanf("%d", &n);
  scanf("%d %lld", &L, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    if (c[i] > S) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    G[x].push_back(i);
  }
  dfs(1);
  printf("%d", ans + (int)s[1].size());
}
