#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const int N = 1e5 + 10;
vector<ll> G[N];
vector<P> ord;
;
vector<P> f[N];
ll n, L, S;
ll w[N];
int cnt = 0;
void dfs(int cur) {
  for (auto u : G[cur]) dfs(u);
  ord.push_back({++cnt, cur});
}
int main() {
  cin >> n >> L >> S;
  for (ll i = 1; i <= n; i++) {
    cin >> w[i];
    if (w[i] > S) return puts("-1"), 0;
  }
  for (ll i = 2; i <= n; i++) {
    ll p;
    cin >> p;
    G[p].push_back(i);
  }
  dfs(1);
  sort(ord.begin(), ord.end());
  ll ans = 0;
  for (int i = 0; i < ord.size(); ++i) {
    auto tmp = ord[i];
    for (auto j : G[tmp.second]) {
      for (auto p : f[j])
        if (p.first + 1 <= L && p.second + w[tmp.second] <= S)
          f[tmp.second].push_back({p.first + 1, p.second + w[tmp.second]});
    }
    if (f[tmp.second].empty())
      f[tmp.second].push_back({1, w[tmp.second]}), ++ans;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const int N = 1e5 + 10;
vector<ll> G[N];
vector<P> ord;
;
vector<P> f[N];
ll n, L, S;
ll w[N];
int cnt = 0;
void dfs(int cur) {
  for (auto u : G[cur]) dfs(u);
  ord.push_back({++cnt, cur});
}
int main() {
  cin >> n >> L >> S;
  for (ll i = 1; i <= n; i++) {
    cin >> w[i];
    if (w[i] > S) return puts("-1"), 0;
  }
  for (ll i = 2; i <= n; i++) {
    ll p;
    cin >> p;
    G[p].push_back(i);
  }
  dfs(1);
  sort(ord.begin(), ord.end());
  ll ans = 0;
  for (int i = 0; i < ord.size(); ++i) {
    auto tmp = ord[i];
    for (auto j : G[tmp.second]) {
      for (auto p : f[j])
        if (p.first + 1 <= L && p.second + w[tmp.second] <= S)
          f[tmp.second].push_back({p.first + 1, p.second + w[tmp.second]});
    }
    if (f[tmp.second].empty())
      f[tmp.second].push_back({1, w[tmp.second]}), ++ans;
  }
  cout << ans << endl;
  return 0;
}

```