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, L, fa[100010], vis[100010] = {0}, ans = 0, a[100010];
long long S;
vector<int> g[100010];
priority_queue<pair<int, int> > q;
void fail() {
  printf("-1\n");
  exit(0);
}
void dfs(int u, int dep) {
  q.push({dep, u});
  for (int i = 0; i < g[u].size(); ++i) dfs(g[u][i], dep + 1);
}
int main() {
  scanf("%d%d%I64d", &n, &L, &S);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] > S) fail();
  }
  for (int i = 2; i <= n; ++i) {
    scanf("%d", &fa[i]);
    g[fa[i]].push_back(i);
  }
  dfs(1, 1);
  pair<int, int> u;
  int cnt, cu;
  long long cur;
  while (!q.empty()) {
    u = q.top();
    q.pop();
    if (vis[u.second]) continue;
    ++ans;
    cu = u.second;
    cnt = 1;
    cur = a[cu];
    vis[cu] = 1;
    while (fa[cu] && cnt < L && cur + a[fa[cu]] <= S) {
      cu = fa[cu];
      cur += a[cu];
      ++cnt;
      vis[cu] = 1;
    }
  }
  printf("%d\n", ans);
  return 0;
}
