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 maxn = 1e5 + 233;
vector<int> g[maxn];
int p[maxn][20], d[maxn], l[maxn], h[maxn];
long long f[maxn][20], S;
long long s[maxn];
int ans = 0, n, L;
int cal(int x) {
  long long ss = s[x];
  int ll = l[x];
  for (int j = 19; j >= 0; j--)
    if (p[x][j]) {
      if (f[x][j] > ss) continue;
      if ((1 << j) > ll) continue;
      ss -= f[x][j];
      ll -= 1 << j;
      x = p[x][j];
    }
  if (ll && f[x][0] <= ss) x = p[x][0];
  return d[x];
}
int main() {
  cin >> n >> L >> S;
  memset(p, 0, sizeof(p));
  d[1] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f[i][0]);
    if (f[i][0] > S) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    scanf("%d", &p[i][0]);
    g[p[i][0]].push_back(i);
    d[i] = d[p[i][0]] + 1;
  }
  for (int j = 1; j <= 19; j++)
    for (int i = 1; i <= n; i++)
      if (p[i][j - 1] && p[p[i][j - 1]][j - 1]) {
        int t = p[i][j - 1];
        p[i][j] = p[t][j - 1];
        f[i][j] = f[i][j - 1] + f[t][j - 1];
      }
  for (int i = n; i >= 1; i--) {
    int mini = d[i];
    for (int j = 0; j < g[i].size(); j++) {
      int y = g[i][j];
      if (mini > h[y]) {
        mini = h[y];
        l[i] = l[y] - 1;
        s[i] = s[y] - f[i][0];
      }
    }
    if (mini == d[i]) {
      ans++;
      l[i] = L;
      s[i] = S;
    }
    h[i] = cal(i);
  }
  cout << ans << endl;
  return 0;
}
