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;
long long n, l, s, w[500000], fa[500000], vis[500000], d[500000], head[500000],
    ver[500000];
long long next1[500000], tot;
void add(long long x, long long y) {
  ver[++tot] = y;
  next1[tot] = head[x];
  head[x] = tot;
}
void bfs() {
  long long first = 1, last = 1;
  vis[1] = 1;
  d[1] = 1;
  while (first <= last) {
    long long x = d[first++];
    for (long long i = head[x]; i; i = next1[i]) {
      long long y = ver[i];
      if (!vis[y]) {
        vis[y] = 1;
        d[++last] = y;
      }
    }
  }
}
int main() {
  cin >> n >> l >> s;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    if (w[i] > s) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (long long i = 2; i <= n; i++) {
    scanf("%lld", &fa[i]);
    add(i, fa[i]);
    add(fa[i], i);
  }
  bfs();
  long long ans = 0, now = 0, len = 0;
  tot = 0;
  memset(vis, 0, sizeof(vis));
  for (long long i = n; i >= 1; i--) {
    long long x = d[i];
    if (!vis[x]) {
      ans++;
      len = 1;
      tot = 0;
      now = x;
      while (now != 0 && len <= l && tot + w[now] <= s) {
        vis[now] = 1;
        len++;
        tot += w[now];
        now = fa[now];
      }
    }
  }
  cout << ans << endl;
}
