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 MAXN = 1e5 + 10;
vector<int> gr[MAXN];
long long w[MAXN];
int t[21][MAXN];
long long sp[MAXN];
int h[MAXN];
pair<int, int> dp[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, l;
  long long s;
  cin >> n >> l >> s;
  dp[0].first = 1e9;
  for (int i = 1; i <= n; ++i) {
    cin >> w[i];
    if (w[i] > s) {
      cout << -1;
      return 0;
    }
    dp[i] = {0, 0};
  }
  for (int i = 2; i <= n; ++i) {
    cin >> t[0][i];
    gr[t[0][i]].emplace_back(i);
  }
  h[1] = 1;
  for (int node = 1; node <= n; ++node) {
    sp[node] += w[node];
    for (int i = 1; i < 20; ++i) {
      t[i][node] = t[i - 1][t[i - 1][node]];
    }
    for (auto &x : gr[node]) {
      h[x] = h[node] + 1;
      sp[x] += sp[node];
    }
  }
  for (int node = n; node > 0; --node) {
    int fiu = 0;
    for (auto &x : gr[node]) {
      if (dp[x].first < dp[fiu].first) {
        fiu = x;
      }
    }
    if (fiu && dp[fiu].first <= h[node]) {
      dp[node].first = dp[fiu].first;
      dp[node].second = 0;
      for (auto &x : gr[node]) dp[node].second += dp[x].second;
    } else {
      int best = node;
      long long sum = w[node];
      int lungime = 1;
      for (int step = 18; step >= 0; --step) {
        if (lungime + (1 << step) > l) continue;
        if (t[step][best] == 0) continue;
        if (sum + sp[t[0][best]] - sp[t[0][t[step][best]]] > s) continue;
        sum += sp[t[0][best]] - sp[t[0][t[step][best]]];
        best = t[step][best];
        lungime += (1 << step);
      }
      dp[node].first = h[best];
      dp[node].second = 1;
      for (auto &x : gr[node]) {
        dp[node].second += dp[x].second;
      }
    }
  }
  cout << dp[1].second << '\n';
  return 0;
}
