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 N = 100005;
int n, log2n, L;
int ft[N][50], de[N], l[N], f[N], g[N];
long long sum[N][50], w[N], S;
vector<int> a[N];
int savelog(int x) { return round(log2(x)); }
void dfs(int u, int fa) {
  ft[u][0] = fa;
  sum[u][0] = w[u];
  for (int _i = 0; _i < a[u].size(); _i++) {
    int v = a[u][_i];
    if (v == fa) continue;
    de[v] = de[u] + 1;
    dfs(v, u);
  }
}
void dp(int u, int fa) {
  bool bo = 1, flag = 0;
  for (int _i = 0; _i < a[u].size(); _i++) {
    int v = a[u][_i];
    if (v == fa) continue;
    dp(v, u);
    if (g[v] > 1) g[u] = max(g[v] - 1, g[u]), flag = 1;
    f[u] += f[v];
    bo = 0;
  }
  if (!flag) f[u]++, g[u] = l[u];
  if (bo) g[u] = l[u], f[u] = 1;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> L >> S;
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    if (w[i] > S) {
      cout << -1;
      return 0;
    }
  }
  log2n = savelog(n);
  for (int i = 2; i <= n; i++) {
    int u;
    cin >> u;
    a[u].push_back(i);
    a[i].push_back(u);
  }
  dfs(1, 0);
  for (int j = 1; j <= log2n; j++)
    for (int i = 1; i <= n; i++)
      ft[i][j] = ft[ft[i][j - 1]][j - 1],
      sum[i][j] = sum[i][j - 1] + sum[ft[i][j - 1]][j - 1];
  for (int i = 1; i <= n; i++) {
    l[i] = 0;
    long long cnt = 0;
    int k = min(de[i] + 1, L);
    int logk = savelog(k);
    for (int j = logk; j >= 0; j--)
      if ((k & (1 << j)) && cnt + sum[i][j] <= S)
        cnt += sum[i][j], l[i] += (1 << j);
  }
  dp(1, 0);
  cout << f[1];
  return 0;
}
