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;
vector<int> G[100010];
int a[25][100010];
int b[100010];
int c[100010];
int e[100010];
long long Q[100010];
int f[100010];
long long n, l, s;
void dfs1(int u, int pre) {
  Q[u] = Q[pre] + e[u];
  b[u] = b[pre] + 1;
  c[u] = u;
  int p = l;
  for (int k = 20; k >= 0; k--) {
    int f = a[k][c[u]];
    if ((1 << k) >= p || f == -1) continue;
    if (Q[u] - Q[f] + e[f] > s) continue;
    p -= (1 << k);
    c[u] = f;
  }
  for (int i = 0; i < G[u].size(); i++) dfs1(G[u][i], u);
}
int ans = 0;
void dfs2(int u) {
  int z = -1;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    dfs2(v);
    if (f[v] == v) continue;
    if (z == -1 || b[z] > b[f[v]]) z = f[v];
  }
  if (z == -1) {
    z = c[u];
    ans++;
  }
  f[u] = z;
}
int main() {
  scanf("%lld%lld%lld", &n, &l, &s);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &e[i]);
    if (e[i] > s) {
      cout << -1 << endl;
      return 0;
    }
  }
  int t;
  for (int i = 2; i <= n; i++) {
    scanf("%d", &t);
    a[0][i] = t;
    G[t].push_back(i);
  }
  a[0][1] = -1;
  for (int k = 0; k < 20; k++)
    for (int i = 1; i <= n; i++)
      if (a[k][i] < 0)
        a[k + 1][i] = -1;
      else
        a[k + 1][i] = a[k][a[k][i]];
  dfs1(1, 0);
  dfs2(1);
  cout << ans << endl;
  return 0;
}
