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;
const int csize = 2;
int k, m;
long long val[N];
int par[N], ancestor[N][18], top[N], in[N];
long long ancestor_val[N][18];
vector<int> edge[N];
int dep[N];
void dfs(int x) {
  for (int to : edge[x]) {
    dep[to] = dep[x] + 1;
    dfs(to);
  }
}
void work_on_top(int n, int L, long long S) {
  dep[1] = 1;
  dfs(1);
  for (int i = 1; i <= n; i++) {
    ancestor[i][0] = par[i];
    ancestor_val[i][0] = val[par[i]];
  }
  for (int j = 1; j < 18; j++)
    for (int i = 1; i <= n; i++) {
      if (i + (1 << j) > n) break;
      ancestor[i][j] = ancestor[ancestor[i][j - 1]][j - 1];
      ancestor_val[i][j] =
          ancestor_val[i][j - 1] + ancestor_val[ancestor[i][j - 1]][j - 1];
    }
  for (int i = 1; i <= n; i++) {
    long long num_limit = L - 1, val_limit = S - val[i];
    int cur = 17, x = i;
    while (cur >= 0) {
      if (ancestor[x][cur] == 0 || (1 << cur) > num_limit ||
          ancestor_val[x][cur] > val_limit)
        cur--;
      else {
        val_limit -= ancestor_val[x][cur];
        num_limit -= (1 << cur);
        x = ancestor[x][cur];
      }
    }
    top[i] = x;
  }
}
bool inque[N], vis[N];
priority_queue<pair<int, int> > pq;
void work(int n, int L, long long S) {
  for (int i = 1; i <= n; i++)
    if (!in[i]) {
      pq.push(make_pair(dep[i], i));
      inque[i] = true;
    }
  int ans = 0, cur;
  while (pq.size()) {
    int u = pq.top().second;
    pq.pop();
    if (vis[u]) continue;
    int top_now = top[u], now = u, far = par[u];
    while (dep[far] > dep[top_now]) {
      vis[now] = true;
      par[now] = top_now;
      now = far;
      far = par[now];
    }
    vis[now] = 1;
    vis[top_now] = 1;
    if (now != top_now) par[now] = top_now;
    ans++;
    if (par[top_now] && !vis[par[top_now]] && !inque[par[top_now]]) {
      pq.push(make_pair(dep[par[top_now]], par[top_now]));
      inque[par[top_now]] = 1;
    }
  }
  printf("%d\n", ans);
}
int main() {
  int n, L;
  long long S, max_val = 0;
  scanf("%d %d %lld", &n, &L, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &val[i]);
    max_val = max(max_val, val[i]);
  }
  par[1] = 0;
  for (int i = 2; i <= n; i++) {
    scanf("%d", &par[i]), in[par[i]]++;
    edge[par[i]].push_back(i);
  }
  if (max_val > S) {
    puts("-1");
    return 0;
  }
  work_on_top(n, L, S);
  work(n, L, S);
  return 0;
}
