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;
class Solution {
 public:
  int minimize(vector<vector<int>>& graph, vector<int>& ws, int n, int l,
               long long s) {
    vector<pair<int, long long>> memo(n);
    _graph = graph;
    _ws = ws;
    _l = l;
    _s = s;
    for (auto w : ws) {
      if (w > s) {
        return -1;
      }
    }
    vector<long long> pv;
    pv.push_back(0);
    return dfs(0, pv, memo);
  }

 private:
  vector<vector<int>> _graph;
  vector<int> _ws;
  int _l;
  long long _s;
  bool debug;
  int dfs(int u, vector<long long>& pv, vector<pair<int, long long>>& memo) {
    int res = 1;
    pv.push_back(pv.back() + _ws[u]);
    for (auto v : _graph[u]) {
      res += dfs(v, pv, memo);
    }
    int maxLen = -1;
    int vv = -1;
    memo[u].first = 1;
    memo[u].second = _ws[u];
    for (auto v : _graph[u]) {
      if (1 + memo[v].first > _l) {
        continue;
      }
      long long w = _ws[u] + memo[v].second;
      if (w > _s) {
        continue;
      }
      long long t = pv.back() + memo[v].second - _s;
      auto it = lower_bound(pv.begin(), pv.end(), t);
      int j = it - pv.begin();
      int len = pv.size() - 1 - j;
      len = min(len, _l - memo[v].first);
      if (len > maxLen) {
        maxLen = len;
        vv = v;
      }
    }
    if (vv != -1) {
      --res;
      memo[u].first = 1 + memo[vv].first;
      memo[u].second = _ws[u] + memo[vv].second;
    }
    pv.pop_back();
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, l;
  long long s;
  cin >> n >> l >> s;
  vector<int> ws(n);
  for (int i = 0; i < n; ++i) {
    cin >> ws[i];
  }
  vector<vector<int>> graph(n);
  for (int i = 1, p = -1; i <= n - 1; ++i) {
    cin >> p;
    graph[p - 1].push_back(i);
  }
  Solution sol;
  cout << sol.minimize(graph, ws, n, l, s) << endl;
}
