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;
using ll = long long;
int n, L;
ll S;
vector<int> a;
vector<int> hi;
vector<int> dist;
vector<vector<int>> g;
vector<int> stk;
ll total_sum;
vector<ll> fa;
ll query(int x) {
  ll y = total_sum;
  for (int i = x - 1; i > 0; i -= (i & -i)) y -= fa[i];
  return y;
}
void add(int x, ll v) {
  for (int i = x; i <= n; i += (i & -i)) fa[i] += v;
}
void dfs(int u, int p) {
  stk.emplace_back(u);
  dist[u] = dist[p] + 1;
  total_sum += a[u];
  add(dist[u], a[u]);
  {
    int r = dist[u];
    int l = max(1, r - L + 1);
    int d = r;
    while (l <= r) {
      int m = (l + r) / 2;
      ll x = query(m);
      if (x <= S) {
        d = m;
        r = m - 1;
      } else {
        l = m + 1;
      }
    }
    int node = stk[d - 1];
    hi[u] = node;
  }
  for (auto v : g[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
  total_sum -= a[u];
  add(dist[u], -a[u]);
  stk.pop_back();
}
int ans = 0;
int solve(int u, int p) {
  int node = 0;
  for (auto v : g[u]) {
    if (v == p) continue;
    int x = solve(v, u);
    if (dist[x] <= dist[u] && (node == 0 || dist[x] < dist[node])) node = x;
  }
  if (node == 0) {
    ++ans;
    return hi[u];
  } else {
    return node;
  }
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cin >> n >> L >> S;
  a.resize(n + 1);
  g.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] > S) return cout << -1, 0;
  }
  for (int i = 2; i <= n; ++i) {
    int p;
    cin >> p;
    g[i].emplace_back(p);
    g[p].emplace_back(i);
  }
  dist.resize(n + 1);
  hi.resize(n + 1);
  fa.resize(n + 10);
  dfs(1, 0);
  solve(1, 0);
  cout << ans;
}
