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;
int n, L;
int p[100003][24];
long long S;
long long W[100003];
long long sum[100003][24];
vector<int> vec[100003];
void build(int u) {
  for (int i = 1; i < 20; i++) {
    if (p[u][i - 1] == -1) break;
    p[u][i] = p[p[u][i - 1]][i - 1];
    sum[u][i] = sum[u][i - 1] + sum[p[u][i - 1]][i - 1] - W[p[u][i - 1]];
  }
  for (int v : vec[u]) build(v);
}
pair<long long, int> best(int u, pair<long long, int> a,
                          pair<long long, int> b) {
  if (a.first == -1)
    return b;
  else {
    int ua = u;
    int ub = u;
    int ja = 0, jb = 0;
    long long sa = a.first, sb = b.first;
    for (int i = 19; i >= 0; i--) {
      if (p[ua][i] != -1 && sa + sum[ua][i] - W[ua] <= S) {
        sa += sum[ua][i] - W[ua];
        ja += (1 << i);
        ua = p[ua][i];
      }
      if (p[ub][i] != -1 && sb + sum[ub][i] - W[ub] <= S) {
        sb += sum[ub][i] - W[ub];
        jb += (1 << i);
        ub = p[ub][i];
      }
      ja = min(ja, L - a.second);
      jb = min(jb, L - b.second);
    }
    if (ja > jb)
      return a;
    else
      return b;
  }
}
int res;
pair<long long, int> dp[100003];
void dfs(int u) {
  dp[u] = {-1, -1};
  for (int v : vec[u]) {
    dfs(v);
    int l = dp[v].second;
    long long w = dp[v].first;
    if (w + W[u] <= S && l != L) {
      dp[u] = best(u, dp[u], {w + W[v], l + 1});
    }
  }
  if (dp[u].first == -1) {
    res++;
    dp[u] = {W[u], 1};
  }
}
int main() {
  memset(W, -1, sizeof W);
  memset(p, -1, sizeof p);
  scanf("%d %d %lld", &n, &L, &S);
  bool bol = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &W[i]);
    if (W[i] > S) bol = 1;
  }
  for (int i = 2; i <= n; i++) {
    scanf("%d", &p[i][0]);
    sum[i][0] = W[i] + W[p[i][0]];
    vec[p[i][0]].push_back(i);
  }
  if (bol) {
    printf("-1\n");
    return 0;
  }
  build(1);
  dfs(1);
  printf("%d\n", res);
}
