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>
const int MAX_N = 1e5 + 5;
const int LOG = 20;
int n, l, p[MAX_N][LOG], h[MAX_N], ans;
long long s, w[MAX_N];
std::vector<int> adj[MAX_N];
int solve(int x) {
  w[x] += w[p[x][0]];
  int min = h[x] = h[p[x][0]] + 1;
  for (int y : adj[x]) min = std::min(min, solve(y));
  if (min == h[x]) {
    int cnt = 0, ind = x;
    for (int step = LOG - 1; step >= 0; --step)
      if (cnt + (1 << step) <= l && w[x] - w[p[ind][step]] <= s) {
        cnt += 1 << step;
        ind = p[ind][step];
      }
    min = h[ind];
    ++ans;
  }
  return min;
}
int main() {
  scanf("%d%d%lld", &n, &l, &s);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", w + i);
    if (w[i] > s) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 2; i <= n; ++i) {
    scanf("%d", p[i]);
    adj[p[i][0]].push_back(i);
  }
  for (int k = 1; k < LOG; ++k)
    for (int i = 1; i <= n; ++i) p[i][k] = p[p[i][k - 1]][k - 1];
  solve(1);
  printf("%d\n", ans);
  return 0;
}
