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;
vector<long long> v[100005];
long long a[100005], vs[100005], w[100005], p[100005], pr[100005][22],
    sum[100005][22];
long long node[100005], sumnode[100005], mul[30], ans = 0, d[100005];
long long checknode(long long summ, long long n) {
  long long i, j, k;
  if (summ <= 0) return 0;
  for (i = 0; i < 20; i++) {
    if (sum[n][i] > summ) {
      if (i == 0)
        return 0;
      else
        return mul[i - 1] + checknode(summ - sum[n][i - 1], pr[n][i - 1]);
    }
    if (pr[n][i] == 0) return d[n] + 1;
  }
}
long long dfs(long long n, long long s, long long l) {
  long long i, j = 0, m, k = 0, kk = -1;
  m = v[n].size();
  for (i = 0; i < m; i++) {
    dfs(v[n][i], s, l);
  }
  for (i = 0; i < m; i++) {
    j = checknode(s - sumnode[v[n][i]], n);
    j = min(j, l - node[v[n][i]]);
    if (j > k) {
      k = j;
      kk = i;
    }
    if (sumnode[v[n][i]] > s) ans--;
  }
  if (m == 0 || kk == -1) {
    sumnode[n] = w[n];
    node[n] = 1;
    ans += m;
  } else {
    sumnode[n] = sumnode[v[n][kk]] + w[n];
    node[n] = node[v[n][kk]] + 1;
    ans += (m - 1);
  }
  return 1;
}
long long dfs1(long long n) {
  long long i, j, k, m;
  for (i = 0; i < 20; i++) {
    if (i == 0)
      pr[n][i] = p[n];
    else
      pr[n][i] = pr[pr[n][i - 1]][i - 1];
  }
  for (i = 0; i < 20; i++) {
    if (i == 0)
      sum[n][i] = w[n];
    else
      sum[n][i] = sum[pr[n][i - 1]][i - 1] + sum[n][i - 1];
  }
  for (i = 0; i < v[n].size(); i++) {
    d[v[n][i]] = d[n] + 1;
    dfs1(v[n][i]);
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long j, n, m, k = 0, l = 0, s, i, pp = 0, kk = 0;
  mul[0] = 1;
  for (i = 1; i < 20; i++) mul[i] = mul[i - 1] * 2;
  cin >> n >> l >> s;
  for (i = 1; i <= n; i++) {
    cin >> w[i];
    pp += w[i];
    if (w[i] > s) kk = -1;
  }
  for (i = 2; i <= n; i++) {
    cin >> j;
    v[j].push_back(i);
    p[i] = j;
    if (j != i - 1) k = 1;
  }
  dfs1(1);
  dfs(1, s, l);
  ans == 0 ? ans-- : ans++;
  if (ans == -1) {
    if (pp <= s && n <= l && k == 0) ans += 2;
  }
  if (kk == -1) ans = -1;
  cout << ans << endl;
  return 0;
}
