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;
inline bool isvowel(char c) {
  c = tolower(c);
  if (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u')
    return 1;
  return 0;
}
const double eps = 0.000001;
const long double pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const long long MOD = 1e18 + 9;
const long long INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 6e5 + 123;
const int M = 22;
const int pri = 997;
const int Magic = 2101;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int n, L;
long long second;
vector<int> g[N];
int cnt[N], col[N];
int a[N], cur;
vector<long long> vec;
long long sum[N];
void dfs(int v) {
  long long best = 0;
  for (auto to : g[v]) {
    vec.push_back(vec.back() + a[to]);
    dfs(to);
    vec.pop_back();
    best = max(best, sum[to]);
  }
  if (best >= a[v]) {
    sum[v] = best;
  } else {
    cur++;
    sum[v] = min(second, vec.back() - vec[(int)vec.size() - L - 1]);
  }
  sum[v] -= a[v];
}
int main() {
  cin >> n >> L >> second;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > second) {
      cout << -1;
      exit(0);
    }
  }
  for (long long i = 2; i <= n; i++) {
    int p;
    cin >> p;
    g[p].push_back(i);
  }
  for (int i = 0; i <= L; i++) vec.push_back(0);
  vec.push_back(a[1]);
  dfs(1);
  cout << cur << '\n';
  return 0;
}
