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 SI = 1e5 + 100;
int n, L;
long long S;
long long val[SI];
int MD[SI];
vector<int> G[SI];
map<int, long long> who[SI];
void dfs(int ad) {
  if (G[ad].size() == 0) {
    who[ad].insert(pair<int, int>(1, val[ad]));
    MD[ad] = 1;
    return;
  }
  int tot = 0;
  for (auto it : G[ad]) {
    dfs(it);
    tot += MD[it];
    for (auto ii : who[it]) {
      if (ii.first >= L) break;
      if (ii.second + val[ad] > S) continue;
      if (who[ad].find(ii.first + 1) != who[ad].end())
        who[ad][ii.first + 1] = min(who[ad][ii.first + 1], ii.second + val[ad]);
      else
        who[ad][ii.first + 1] = ii.second + val[ad];
    }
    who[it].clear();
  }
  if (who[ad].size() == 0) {
    tot++;
    MD[ad] = tot;
    who[ad].insert(pair<int, int>(1, val[ad]));
    return;
  }
  MD[ad] = tot;
}
int main() {
  scanf("%d %d %I64d", &n, &L, &S);
  bool cc = true;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &val[i]);
    if (val[i] > S) cc = false;
  }
  for (int i = 2; i <= n; i++) {
    int f;
    scanf("%d", &f);
    G[f].push_back(i);
  }
  if (cc == false) {
    printf("-1");
    return 0;
  }
  dfs(1);
  printf("%d", MD[1]);
  return 0;
}
