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 = 1e16;
const long long MOD = 1e15 + 9;
const long long MX = 1e5 + 9;
int n, len;
long long siz;
long long w[MX];
pair<long long, long long> dp[20][MX];
void build() {
  pair<long long, long long> x;
  for (int i = 1; i < 19; i++) {
    for (int j = 1; j <= n; j++) {
      x = dp[i - 1][j];
      dp[i][j] = dp[i - 1][x.first];
      dp[i][j].second += x.second;
    }
  }
}
long long res;
vector<int> adj[MX];
int check(int ver) {
  int ret = 0;
  long long sum = 0;
  pair<long long, long long> x;
  for (int i = 18; ver != -1 && i >= 0; i--) {
    x = dp[i][ver];
    if (x.first != 0 && sum + x.second <= siz) {
      ver = x.first;
      sum += x.second;
      ret += (1 << i);
    }
  }
  if (ver == 1 && dp[0][1].second + sum <= siz) ret++;
  ret = min(ret, len);
  return ret;
}
int DFS(int ver) {
  int ret = 0;
  for (auto i : adj[ver]) {
    ret = max(ret, DFS(i));
  }
  if (ret == 0) {
    ret = check(ver);
    if (ret == 0) {
      cout << -1;
      exit(0);
    }
    res++;
  }
  ret--;
  return ret;
}
int main() {
  cin >> n >> len >> siz;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
  }
  dp[0][1] = {-1, w[1]};
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    adj[x].push_back(i);
    dp[0][i] = {x, w[i]};
  }
  build();
  DFS(1);
  cout << res;
  return 0;
}
