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 long long N = 100004;
long long n, L, S, w[N], bit[N], id[N], kq = 0;
vector<long long> a[N];
void update(long long x, long long val) {
  for (long long i = x; i <= n; i += (i & -i)) bit[i] += val;
}
long long get(long long x) {
  if (x == 0) return 0;
  long long ans = 0;
  for (long long i = x; i > 0; i -= (i & -i)) ans += bit[i];
  return ans;
}
void dfs1(long long x, long long hi) {
  update(hi, w[x]);
  for (long long i = 0; i < a[x].size(); i++) {
    dfs1(a[x][i], hi + 1);
  }
  long long keep = get(hi);
  long long l = 1, r = hi, res = 0;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (keep - get(mid - 1) <= S)
      res = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  id[x] = min(hi - res + 1, L);
  update(hi, -w[x]);
}
long long dfs(long long x) {
  long long keep = 0;
  for (long long i = 0; i < a[x].size(); i++) {
    keep = max(keep, dfs(a[x][i]));
  }
  if (keep == 0) {
    kq++;
    return id[x] - 1;
  }
  return keep - 1;
}
int main() {
  scanf("%lld%lld%lld", &n, &L, &S);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    if (w[i] > S) {
      cout << "-1";
      return 0;
    }
  }
  for (long long i = 1; i < n; i++) {
    long long u;
    scanf("%lld", &u);
    a[u].push_back(i + 1);
  }
  dfs1(1, 1);
  dfs(1);
  cout << kq;
}
