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 = 100010;
int get() {
  int x;
  bool f = 0;
  char c;
  while (!isdigit(c = getchar()))
    if (c == '-') f = 1;
  x = c - 48;
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + c - 48;
  if (f)
    return -x;
  else
    return x;
}
int N, M, Ans, A[MAXN], depth[MAXN], fa[MAXN][20];
long long K, dis[MAXN];
vector<int> To[MAXN];
int cal(int x) {
  int sum1 = 1;
  long long sum2 = A[x];
  for (int i = 17; i >= 0; --i)
    if (fa[x][i]) {
      if (sum1 + (1 << i) > M) continue;
      if (sum2 + dis[fa[x][0]] - dis[fa[fa[x][i]][0]] <= K) {
        sum1 += (1 << i);
        sum2 += dis[fa[x][0]] - dis[fa[fa[x][i]][0]];
        x = fa[x][i];
      }
    }
  return depth[x];
}
int dfs(int x) {
  int to;
  int Min = depth[x] + 1;
  for (int i = 0; i < To[x].size(); ++i) {
    to = To[x][i];
    depth[to] = depth[x] + 1;
    dis[to] = dis[x] + A[to];
    Min = min(Min, dfs(to));
  }
  if (Min == depth[x] + 1) {
    ++Ans;
    return cal(x);
  } else
    return Min;
}
int main() {
  N = get();
  M = get();
  cin >> K;
  for (int i = 1; i <= N; ++i) {
    A[i] = get();
    if (A[i] > K) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 2; i <= N; ++i) fa[i][0] = get(), To[fa[i][0]].push_back(i);
  for (int j = 1; j <= 17; ++j)
    for (int i = 1; i <= N; ++i) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  depth[1] = 1;
  dis[1] = A[1];
  dfs(1);
  cout << Ans << endl;
  return 0;
}
