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 = 100005;
int a[N];
vector<int> G[N];
int f[N][20], e[N];
long long d[N][20];
int deep[N];
int ans = 0;
void dfs(int u) {
  int now = u;
  for (auto v : G[u]) {
    deep[v] = deep[u] + 1;
    dfs(v);
    if (deep[e[v]] < deep[now]) {
      now = e[v];
    }
  }
  if (now == u) {
    ans++;
  } else {
    e[u] = now;
  }
}
int main() {
  int n, m;
  long long r;
  cin >> n >> m >> r;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] > r) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    int fa;
    scanf("%d", &fa);
    G[fa].push_back(i);
    f[i][0] = fa;
    d[i][0] = a[i];
  }
  int wc = 0;
  while (1 << (wc + 1) <= n) {
    wc++;
  }
  f[1][0] = 0;
  d[1][0] = a[1];
  for (int j = 1; j <= wc; j++) {
    for (int i = 1; i <= n; i++) {
      f[i][j] = f[f[i][j - 1]][j - 1];
      d[i][j] = d[i][j - 1] + d[f[i][j - 1]][j - 1];
    }
  }
  for (int i = 1; i <= n; i++) {
    int k = 0, sum = 0, u = i;
    for (int j = wc; j >= 0 && u != 0; j--) {
      if (k + (1 << j) <= m && sum + (d[u][j]) <= r) {
        k += 1 << j;
        sum += d[u][j];
        u = f[u][j];
      }
    }
    e[i] = u;
  }
  deep[1] = 1;
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
