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 long long INF = 1e9 + 9;
const long long MOD = 1e9 + 7;
const long long LINF = (long long)1e18 + 3;
long long n, l, s;
const long long SIZE = 1e6 + 6;
const long long LOGN = 30;
vector<int> adj[SIZE];
long long waga[SIZE];
long long pref[SIZE];
int parent[SIZE][LOGN];
pair<long long, pair<long long, long long> > dp[SIZE];
void dfs1(long long v, long long prev, long long dep, long long suma) {
  suma += waga[v];
  pref[v] = suma;
  parent[v][0] = prev;
  for (auto it : adj[v]) {
    if (it != prev) dfs1(it, v, dep + 1, suma);
  }
}
void dfs2(long long v, long long prev) {
  for (auto it : adj[v]) {
    if (it != prev) dfs2(it, v);
  }
  if (adj[v].size() == 1 && v != 1) {
    dp[v] = {waga[v], {1, 1}};
  } else {
    bool czy = false;
    long long tmpsum = 0;
    for (auto it : adj[v]) {
      if (it != prev) tmpsum += dp[it].second.second;
      if (it != prev && dp[it].first + waga[v] <= s &&
          dp[it].second.first + 1 <= l)
        czy = true;
    }
    if (czy == false || l == 1) {
      dp[v] = {waga[v], {1, tmpsum + 1}};
      return;
    } else {
      pair<int, long long> best;
      best = {-1, -1};
      for (auto it : adj[v]) {
        if (it == prev || dp[it].first + waga[v] > s ||
            dp[it].second.first == l)
          continue;
        long long l2 = 0, r = n;
        while (l2 + 1 < r) {
          long long mid = (l2 + r) / 2;
          long long vertex = parent[v][0];
          for (long long j = 0; j < LOGN; j++) {
            if (mid & (1 << j)) vertex = parent[vertex][j];
          }
          if (pref[v] - pref[parent[vertex][0]] + dp[it].first <= s)
            l2 = mid;
          else
            r = mid;
        }
        l2 = min(l2, l - dp[it].second.first);
        if (l2 > best.second) {
          best.second = l2;
          best.first = it;
        }
      }
      dp[v] = {waga[v] + dp[best.first].first,
               {dp[best.first].second.first + 1, tmpsum}};
      return;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> s;
  for (long long i = 1; i <= n; i++) {
    cin >> waga[i];
    if (waga[i] > s) {
      cout << -1;
      return 0;
    }
  }
  for (long long i = 2; i <= n; i++) {
    long long x;
    cin >> x;
    adj[i].push_back(x);
    adj[x].push_back(i);
  }
  dfs1(1, 0, 0, 0);
  for (long long i = 1; i < LOGN; i++)
    for (long long j = 1; j <= n; j++)
      parent[j][i] = parent[parent[j][i - 1]][i - 1];
  dfs2(1, -1);
  cout << dp[1].second.second;
}
