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, ll, w[100010], num, k[100010], fa[100010], head[100010], f[100010],
    ans[100010], dp[100010];
long long s, dis[100010];
vector<int> v;
struct xx {
  int next, to;
} way[100010];
void add(int from, int to) {
  way[++num].next = head[from];
  way[num].to = to;
  head[from] = num;
}
void dfs(int x) {
  v.push_back(x);
  dis[x] = dis[fa[x]] + w[x];
  int l = max(0, (int)v.size() - ll), r = v.size() - 1;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (dis[x] - dis[fa[v[mid]]] <= s)
      r = mid;
    else
      l = mid + 1;
  }
  k[x] = v.size() - 1 - l;
  for (int i = head[x]; i; i = way[i].next) {
    int y = way[i].to;
    dfs(y);
  }
  v.pop_back();
}
void dfss(int x) {
  dp[x] = -1;
  for (int i = head[x]; i; i = way[i].next) {
    int y = way[i].to;
    dfss(y);
    dp[x] = max(dp[x], dp[y] - 1);
    ans[x] += ans[y];
  }
  if (dp[x] == -1) {
    dp[x] = k[x];
    ans[x]++;
  }
}
int main() {
  scanf("%d%d%lld", &n, &ll, &s);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    if (w[i] > s) {
      cout << "-1";
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    int g;
    scanf("%d", &g);
    fa[i] = g;
    add(g, i);
  }
  dfs(1);
  dfss(1);
  printf("%d", ans[1]);
  return 0;
}
