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, p;
long long S;
vector<int> g[100005];
int w[100005];
int ans = 0;
int d = 0;
int dp1[100005];
long long dp2[100005];
long long ft[100005];
void update(int i, long long v) {
  for (; i <= 100000; i += (i & -i)) {
    ft[i] += v;
  }
}
long long query(int i) {
  long long res = 0;
  for (; i > 0; i -= (i & -i)) res += ft[i];
  return res;
}
void dfs(int u) {
  d++;
  update(d, w[u]);
  int cur = 0, idx = -1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    dfs(v);
    int g1 = L - dp1[v], g2 = 0;
    int lo = 1, hi = d;
    while (lo <= hi) {
      int mid = (lo + hi) / 2;
      if (query(d) - query(d - mid) + dp2[v] <= S)
        g2 = mid, lo = mid + 1;
      else
        hi = mid - 1;
    }
    int g = min(g1, g2);
    if (cur < g) {
      cur = g;
      idx = v;
    }
  }
  update(d, -w[u]);
  d--;
  if (cur == 0) {
    ans++;
    dp1[u] = 1;
    dp2[u] = w[u];
  } else {
    dp1[u] = dp1[idx] + 1;
    dp2[u] = dp2[idx] + w[u];
  }
}
int main() {
  scanf("%d%d%lld", &n, &L, &S);
  bool ok = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    if (w[i] > S) ok = 0;
  }
  for (int i = 2; i <= n; i++) scanf("%d", &p), g[p].emplace_back(i);
  if (!ok) return !printf("-1\n");
  dfs(1);
  printf("%d\n", ans);
}
