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 = 100001;
int n, L, x, best[N], b[N], w[N], h[N], p[N][20];
long long S, s[N];
vector<int> v[N];
int res;
void pre(int node, int pnode) {
  s[node] = s[pnode] + w[node];
  p[node][0] = pnode;
  h[node] = h[pnode] + 1;
  for (int cur = 1; (1 << cur) <= n; cur++)
    p[node][cur] = p[p[node][cur - 1]][cur - 1];
  b[node] = node;
  for (int cur = 19; cur >= 0; cur--) {
    if (p[b[node]][cur] == 0) continue;
    if (h[node] - h[p[b[node]][cur]] + 1 > L) continue;
    if (s[node] - s[p[b[node]][cur]] + w[p[b[node]][cur]] > S) continue;
    b[node] = p[b[node]][cur];
  }
  for (auto &i : v[node]) pre(i, node);
}
void dfs(int node) {
  best[node] = -1;
  for (auto &i : v[node]) {
    dfs(i);
    if (best[i] == i) continue;
    if (best[node] == -1 || h[best[i]] < h[best[node]]) best[node] = best[i];
  }
  if (best[node] == -1) {
    res++;
    best[node] = b[node];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> L >> S;
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    if (w[i] > S) return cout << -1 << endl, 0;
  }
  for (int i = 2; i <= n && cin >> x; i++) v[x].push_back(i);
  pre(1, 0);
  dfs(1);
  cout << res << endl;
}
