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 int maxn = 1e5 + 5;
vector<int> G[maxn];
pair<int, long long> work[maxn];
int now, ans;
int n, L, last[maxn], dep[maxn];
long long w[maxn], S;
void Getlast(int u, int now) {
  int l = max(0, now - L), r = now;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (work[mid].second + S >= work[now].second)
      r = mid;
    else
      l = mid + 1;
  }
  last[u] = work[l].first;
}
void dfs(int u) {
  now++;
  work[now].first = u;
  work[now].second = work[now - 1].second + w[u];
  Getlast(u, now);
  for (int i = 0; i < G[u].size(); i++) {
    dep[G[u][i]] = dep[u] + 1;
    dfs(G[u][i]);
  }
  now--;
}
int solve(int u) {
  int mn = dep[u];
  int ret = u;
  for (int i = 0; i < G[u].size(); i++) {
    int llll = solve(G[u][i]);
    if (mn > dep[llll]) {
      mn = dep[llll];
      ret = llll;
    }
  }
  if (ret != u) return ret;
  ans++;
  return last[u];
}
int p[maxn];
int main() {
  scanf("%d%d%lld", &n, &L, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    if (w[i] > S) return puts("-1"), 0;
  }
  for (int i = 2; i <= n; i++) scanf("%d", &p[i]), G[p[i]].push_back(i);
  dep[1] = 1;
  dfs(1);
  solve(1);
  printf("%d\n", ans);
  return 0;
}
