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;
int n, L;
long long S;
const int N = 1e5 + 10;
vector<int> G[N];
long long w[N];
int d[N], ud[N];
long long dat[N * 4];
void update(int x, long long y, int v = 0, int l = 0, int r = n) {
  dat[v] += y;
  if (l + 1 == r) return;
  int mid = (l + r) / 2, chl = v * 2 + 1, chr = v * 2 + 2;
  if (x < mid)
    update(x, y, chl, l, mid);
  else
    update(x, y, chr, mid, r);
}
int query(long long S, int v = 0, int l = 0, int r = n) {
  if (l + 1 == r) return l;
  int mid = (l + r) / 2, chl = v * 2 + 1, chr = v * 2 + 2;
  if (dat[chl] >= S) return query(S, chl, l, mid);
  return query(S - dat[chl], chr, mid, r);
}
void Maxi(int &a, int b) {
  if (b > a) a = b;
}
void dfs(int v = 1, int depth = 0) {
  d[v] = depth;
  update(depth, w[v]);
  long long tot = dat[0];
  if (tot <= S) {
    ud[v] = 0;
  } else {
    ud[v] = query(tot - S) + 1;
  }
  Maxi(ud[v], depth - L + 1);
  for (auto u : G[v]) {
    dfs(u, depth + 1);
  }
  update(depth, -w[v]);
}
int ans = 0;
void Mini(int &a, int b) {
  if (b < a) a = b;
}
int solve(int v = 1) {
  int up = d[v] + 1;
  for (auto u : G[v]) {
    Mini(up, solve(u));
  }
  if (up <= d[v]) return up;
  ans++;
  return ud[v];
}
int main() {
  scanf("%d%d%lld", &n, &L, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", w + i);
    if (w[i] > S) {
      return puts("-1"), 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    G[p].push_back(i);
  }
  dfs();
  solve();
  printf("%d\n", ans);
}
