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;
int w[MAXN], fa[MAXN];
int dfn[MAXN];
vector<int> G[MAXN];
bool vis[MAXN];
void bfs(int s) {
  queue<int> q;
  q.push(s);
  int cnt = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = 1, dfn[++cnt] = u;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (vis[v]) continue;
      q.push(v);
    }
  }
}
int main() {
  int n, l;
  long long s;
  scanf("%d%d%lld", &n, &l, &s);
  int mxx = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    mxx = max(mxx, w[i]);
  }
  for (int i = 2; i <= n; i++) {
    scanf("%d", &fa[i]);
    G[fa[i]].push_back(i);
  }
  if (mxx > s)
    puts("-1");
  else {
    bfs(1);
    memset(vis, 0, sizeof(vis));
    int ans = 0;
    for (int i = n; i >= 1; i--) {
      int u = dfn[i];
      if (vis[u]) continue;
      ans++;
      int nowl = l;
      long long nows = s;
      while (u && nowl > 0 && nows >= w[u])
        nowl--, nows -= w[u], vis[u] = 1, u = fa[u];
    }
    printf("%d\n", ans);
  }
  return 0;
}
