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 N = 1e5 + 5;
int n, l, fa[N][20], f[N], h[N], lim, bin[20];
long long s, g[N], val[N][20];
vector<int> to[N];
void dfs(int u) {
  for (int i = 1; i <= lim; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (int i = 1; i <= lim; i++)
    val[u][i] = val[u][i - 1] + val[fa[u][i - 1]][i - 1];
  for (auto v : to[u]) dfs(v);
}
void dfs2(int u) {
  int maxn = 0, maxid = -1;
  for (auto v : to[u]) {
    dfs2(v);
    f[u] += f[v];
    int x = u, cl = h[v];
    long long cs = g[v];
    for (int i = lim; i >= 0; i--)
      if (fa[x][i] && cl + bin[i] <= l && cs + val[x][i] <= s)
        cl += bin[i], cs += val[x][i], x = fa[x][i];
    if (cl + 1 <= l && cs + val[x][0] <= s) cl++, cs += val[x][0];
    if (cl - h[v] > maxn) maxn = cl - h[v], maxid = v;
  }
  if (maxid != -1)
    h[u] = h[maxid] + 1, g[u] = g[maxid] + val[u][0];
  else
    h[u] = 1, g[u] = val[u][0], ++f[u];
}
int main() {
  scanf("%d%d%lld", &n, &l, &s);
  lim = log(n) / log(2);
  bin[0] = 1;
  for (int i = 1; i <= lim; i++) bin[i] = (bin[i - 1] << 1);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &val[i][0]);
    if (1ll * val[i][0] > s) return printf("-1\n"), 0;
  }
  for (int i = 2; i <= n; i++)
    scanf("%d", &fa[i][0]), to[fa[i][0]].push_back(i);
  dfs(1);
  dfs2(1);
  printf("%d\n", f[1]);
  return 0;
}
