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;
long long s, w[N];
vector<int> g[N];
int f[N];
int par[17][N];
long long sum[17][N];
void dfs(int u) {
  for (int i = 1; i <= 16; ++i) {
    sum[i][u] = sum[i - 1][u] + sum[i - 1][par[i - 1][u]];
    par[i][u] = par[i - 1][par[i - 1][u]];
  }
  for (int v : g[u]) {
    par[0][v] = u;
    sum[0][v] = w[u];
    dfs(v);
  }
}
int get_len(int u, int len, long long sumW) {
  int ret = 0;
  long long curW = 0;
  for (int i = 16; i >= 0; --i)
    if (curW + sum[i][u] <= sumW) {
      curW += sum[i][u];
      u = par[i][u];
      ret += (1 << i);
    }
  return min(ret, len);
}
pair<long long, int> dfs2(int u) {
  f[u] = 1;
  int ver_best = 0;
  int len_best = 0;
  pair<long long, int> ret;
  for (int v : g[u]) {
    pair<long long, int> cur = dfs2(v);
    f[u] += f[v];
    int tmp = get_len(v, l - cur.second, s - cur.first);
    if (tmp == 0) continue;
    if (ver_best == 0 || tmp > len_best) {
      ver_best = v;
      len_best = tmp;
      ret = {cur.first + w[u], cur.second + 1};
    }
  }
  if (ver_best)
    f[u]--;
  else {
    ret = {w[u], 1};
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> l >> s;
  for (int i = 1; i <= n; ++i) {
    cin >> w[i];
    if (w[i] > s) return cout << -1 << '\n', 0;
  }
  for (int i = 2; i <= n; ++i) {
    int u;
    cin >> u;
    g[u].push_back(i);
  }
  par[0][1] = 1;
  dfs(1);
  dfs2(1);
  cout << f[1] << '\n';
}
