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;
vector<int> g[100002];
int par[17][100002];
long long val[17][100002];
int w[100002];
int n, L, p;
long long S;
int ans = 0;
int dfs(int u, int p, int W) {
  par[0][u] = p;
  val[0][u] = W;
  for (int j = 1; j < 17; j++) {
    if (par[j - 1][u] != 0) {
      par[j][u] = par[j - 1][par[j - 1][u]];
      val[j][u] = val[j - 1][u] + val[j - 1][par[j - 1][u]];
    }
  }
  vector<int> cur;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    int res = dfs(v, u, w[u]);
    if (res != 0) cur.emplace_back(res);
  }
  sort(cur.begin(), cur.end());
  reverse(cur.begin(), cur.end());
  if (cur.size() > 0) return cur[0] - 1;
  int v = u, d = 1;
  long long sum = w[u];
  for (int j = 16; j >= 0; j--) {
    if (par[j][v] == 0) continue;
    if (sum + val[j][v] <= S) {
      d += (1 << j);
      sum += val[j][v];
      v = par[j][v];
    }
  }
  d = min(d, L);
  ans++;
  return d - 1;
}
int main() {
  scanf("%d%d%lld", &n, &L, &S);
  bool ok = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    if (w[i] > S) ok = 0;
  }
  for (int i = 2; i <= n; i++)
    scanf("%d", &p), g[p].emplace_back(i), g[i].emplace_back(p);
  if (!ok) {
    printf("-1\n");
    return 0;
  }
  dfs(1, 0, 0);
  printf("%d\n", ans);
}
