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, nowl, ans, now, i, mp[100001], fa[100001];
long long S, a[100001], nows, mx;
int main() {
  scanf("%d%d%lld", &n, &L, &S);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  if (mx > S) {
    puts("-1");
    return 0;
  }
  for (i = 2; i <= n; i++) scanf("%d", &fa[i]);
  for (i = n; i; i--)
    if (!mp[i]) {
      ans++;
      nowl = L;
      nows = S;
      now = i;
      while (now && nowl > 0 && nows >= a[now]) {
        mp[now] = 1;
        nowl--;
        nows -= a[now];
        now = fa[now];
      }
    }
  printf("%d", ans);
}
