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 maxn = 100005;
long long dep[maxn], S;
int par[18][maxn], sum[maxn], w[maxn], n, L;
int jump(int x) {
  int l = L;
  long long s = S;
  for (int i = 17; i >= 0; i--) {
    if ((1 << i) <= l && dep[x] - dep[par[i][x]] <= s)
      s -= dep[x] - dep[par[i][x]], l -= 1 << i, x = par[i][x];
  }
  return x;
}
int main() {
  scanf("%d%d%lld", &n, &L, &S);
  for (int i = 1; i <= n; i++) scanf("%d", w + i);
  dep[1] = w[1];
  for (int i = 2; i <= n; i++) {
    scanf("%d", par[0] + i);
    dep[i] = dep[par[0][i]] + w[i];
  }
  for (int i = 1; i < 18; i++)
    for (int j = 1; j <= n; j++) par[i][j] = par[i - 1][par[i - 1][j]];
  int res = 0;
  for (int i = n; i >= 1; i--) {
    if (!sum[i]) {
      int v = jump(i);
      if (v == i) return puts("-1") * 0;
      --sum[v], ++sum[i];
      ++res;
    }
    sum[par[0][i]] += sum[i];
  }
  printf("%d\n", res);
  return 0;
}
