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 int N = 1e5 + 5;
int n, L;
long long w[N], S, pre[N];
vector<int> g[N];
vector<pair<long long, int>> vec;
int hgt[N], par[N], dp[N], dep[N];
void dfs(int x) {
  vec.push_back(make_pair(pre[par[x]], x));
  pre[x] = pre[par[x]] + w[x];
  for (int it : g[x]) {
    dep[it] = dep[x] + 1;
    dfs(it);
  }
  int y =
      lower_bound(vec.begin(), vec.end(), make_pair(pre[x] - S, -1))->second;
  y = min(L, 1 + dep[x] - dep[y]);
  for (int it : g[x]) {
    hgt[x] = max(hgt[x], hgt[it]);
  }
  if (hgt[x] == 0) {
    dp[x] = 1;
    hgt[x] = y;
  }
  hgt[x]--;
  vec.pop_back();
}
int main() {
  scanf("%d%d%lld", &n, &L, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", w + i);
    if (w[i] > S) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    scanf("%d", par + i);
    g[par[i]].push_back(i);
  }
  dfs(1);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += dp[i];
  }
  printf("%d\n", ans);
}
