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;
long long n, l, s, i, x = 1, w[100005], p[100005], d[100005], f[100005][20],
                      t[100005][20], ans, ss;
vector<long long> g[100005];
void dfs1(long long u) {
  long long v, k = w[u];
  for (long long i = 1; i <= ss; i++)
    f[u][i] = f[v = f[u][i - 1]][i - 1], t[u][i] = t[u][i - 1] + t[v][i - 1];
  for (long long i = 0; i < g[u].size(); i++) {
    v = g[u][i], f[v][0] = u, t[v][0] = w[u], d[v] = d[u] + 1;
    dfs1(v);
    if (d[p[v]] < d[p[u]]) p[u] = p[v];
  }
  if (d[p[u]] > d[u]) {
    v = u;
    for (long long i = ss; i >= 0; i--)
      if (d[u] - d[f[v][i]] <= l && k + t[v][i] <= s) k += t[v][i], v = f[v][i];
    ans++, p[u] = v;
  }
}
int main() {
  scanf("%lld%lld%lld", &n, &l, &s);
  if (!l) ans = -1;
  l--, d[n + 1] = n + 1, t[1][0] = s + 1;
  while (x < n) x <<= 1, ss++;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    if (w[i] > s) ans = -1;
    p[i] = n + 1;
  }
  if (ans == -1) {
    cout << -1 << endl;
    return 0;
  }
  for (i = 2; i <= n; i++) {
    scanf("%lld", &x);
    g[x].push_back(i);
  }
  d[1] = 1, dfs1(1);
  cout << ans << endl;
  return 0;
}
