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 = 1e5 + 101;
long long f[N], ans[N], a[N];
long long head[N], tot, val[N], fa[N][21];
long long n, L, S;
long long deep[N], dis[N], siz[N];
struct Edge {
  long long to, nex;
} edge[N * 2];
void add(long long from, long long to) {
  edge[++tot].nex = head[from];
  edge[tot].to = to;
  head[from] = tot;
}
void dfs(long long now, long long fath) {
  siz[now] = 1;
  dis[now] = dis[fath] + a[now];
  deep[now] = deep[fath] + 1;
  fa[now][0] = fath;
  for (long long i = 1; i <= 20; i++) fa[now][i] = fa[fa[now][i - 1]][i - 1];
  for (long long i = head[now]; i; i = edge[i].nex) {
    long long v = edge[i].to;
    if (v == fath) continue;
    dfs(v, now);
    siz[now] += siz[v];
  }
}
void get_val() {
  for (long long i = 1; i <= n; i++) {
    long long x = i;
    for (long long j = 20; j >= 0; j--) {
      if (fa[x][j] == 0) continue;
      long long y = fa[x][j];
      if (deep[i] - deep[y] + 1 <= L && dis[i] - dis[y] + a[y] <= S) x = y;
    }
    val[i] = deep[i] - deep[x] + 1;
  }
}
void dfs2(long long now, long long fath) {
  if (siz[now] == 1) {
    ans[now] = 1;
    f[now] = val[now] - 1;
    return;
  }
  for (long long i = head[now]; i; i = edge[i].nex) {
    long long v = edge[i].to;
    if (v == fath) continue;
    dfs2(v, now);
    f[now] = max(f[v] - 1, f[now]);
    ans[now] += ans[v];
  }
  if (f[now] == -1) {
    f[now] = val[now] - 1;
    ++ans[now];
  }
}
signed main() {
  memset(f, -1, sizeof(f));
  cin >> n >> L >> S;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > S) {
      puts("-1");
      return 0;
    }
  }
  for (long long i = 2; i <= n; i++) {
    long long x;
    cin >> x;
    add(i, x);
    add(x, i);
  }
  dfs(1, 0);
  get_val();
  dfs2(1, 0);
  cout << ans[1] << endl;
}
