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;
long long w[100001], cost[100001], dist[100001][18], d[100001][18], len[100001];
vector<int> a[100001];
int p[100001], ind[100001];
long long n, L, S;
long long go(long long mid, int y) {
  long long cnt = cost[y];
  if (len[y] + mid > L) return false;
  for (int i = 17; i >= 0; i--) {
    if (d[y][i] == 0 || mid - ((long long)1 << i) < 0) continue;
    cnt += dist[y][i];
    mid -= ((long long)1 << i);
    y = d[y][i];
  }
  if (mid > 0 || cnt > S) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> L >> S;
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    len[i] = 1;
    if (w[i] > S) {
      cout << -1 << '\n';
      return 0;
    }
  }
  if (n == 1) {
    cout << 1 << '\n';
    return 0;
  }
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
    a[p[i]].push_back(i);
    ind[p[i]] += 1;
    d[i][0] = p[i];
    dist[i][0] = w[p[i]];
  }
  for (int i = 1; ((long long)1 << i) <= n; i++) {
    for (int j = 2; j <= n; j++) {
      d[j][i] = d[d[j][i - 1]][i - 1];
      dist[j][i] = dist[d[j][i - 1]][i - 1] + dist[j][i - 1];
    }
  }
  queue<int> q;
  for (int i = 2; i <= n; i++) {
    if (ind[i] == 0) {
      ind[p[i]] -= 1;
      len[i] = 1;
      cost[i] = w[i];
      if (ind[p[i]] == 0) {
        q.push(p[i]);
      }
    }
  }
  int ans = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    int l = 1;
    int r = n;
    while (l <= r) {
      int mid = (l + r) / 2;
      bool f = false;
      for (int i = 0; i < a[x].size(); i++) {
        if (go(mid, a[x][i])) {
          f = true;
          break;
        }
      }
      if (f) {
        l = mid + 1;
      } else
        r = mid - 1;
    }
    if (r == 0) {
      ans += a[x].size();
    } else {
      int idx = -1;
      ans += a[x].size() - 1;
      for (int i = 0; i < a[x].size(); i++) {
        if (go(r, a[x][i])) {
          idx = a[x][i];
          break;
        }
      }
      cost[x] = cost[idx] + w[x];
      len[x] += len[idx];
    }
    ind[p[x]] -= 1;
    if (ind[p[x]] == 0) {
      q.push(p[x]);
    }
  }
  cout << ans + 1 << '\n';
}
