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 maxn = (int)1e5 + 1;
int n, m, w[maxn], fa[maxn], dep[maxn], tot, path[maxn], sum[maxn], ans;
long long lim, s[maxn];
vector<int> e[maxn], rem[maxn];
set<int> sp[2][maxn];
void pfs(int u, int id) {
  path[tot++] = u;
  for (; dep[u] - dep[path[id]] > m || s[u] - s[path[id]] > lim; ++id)
    ;
  ++sum[u];
  --sum[path[id]];
  rem[path[id]].push_back(u);
  for (int v : e[u]) pfs(v, id);
  --tot;
}
void merge(set<int> &lft, set<int> &rht) {
  if (lft.size() < rht.size()) lft.swap(rht);
  for (int v : rht) lft.insert(v);
  set<int>().swap(rht);
}
void dfs(int u) {
  sp[1][u].insert(u);
  for (int v : e[u]) {
    dfs(v);
    sum[u] += sum[v];
    merge(sp[0][u], sp[0][v]);
    merge(sp[1][u], sp[1][v]);
  }
  for (int v : rem[u]) {
    if (sp[0][u].count(v)) sp[0][u].erase(v);
    if (sp[1][u].count(v)) sp[1][u].erase(v);
  }
  if (sp[0][u].empty() && !sp[1][u].empty()) {
    ++ans;
    merge(sp[0][u], sp[1][u]);
  }
}
int main() {
  scanf("%d%d%lld", &n, &m, &lim);
  for (int i = 1; i <= n; ++i) scanf("%d", w + i);
  dep[1] = 1;
  s[1] = w[1];
  for (int i = 2; i <= n; ++i) {
    scanf("%d", fa + i);
    e[fa[i]].push_back(i);
    dep[i] = dep[fa[i]] + 1;
    s[i] = s[fa[i]] + w[i];
  }
  path[tot++] = 0;
  pfs(1, 0);
  dfs(1);
  for (int i = 1; i <= n; ++i)
    if (sum[i] <= 0) {
      puts("-1");
      return 0;
    }
  printf("%d\n", ans);
  return 0;
}
