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 len = 21;
long long n, L, S, l, r;
vector<int> graph[100005], up[100005], w;
vector<long long> sum[100005];
int ans = 0;
void dfs(int v, int p = 0) {
  up[v][0] = p;
  sum[v][0] = w[v];
  for (int i = 1; i <= len; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    sum[v][i] = sum[up[v][i - 1]][i - 1] + sum[v][i - 1];
  }
  for (int i = 0; i < graph[v].size(); ++i) {
    int to = graph[v][i];
    if (to != p) dfs(to, v);
  }
}
long long getMaxDist(int v, long long maxSum) {
  long long cnt = 0;
  for (int i = len; i >= 0; i--)
    if (sum[v][i] <= maxSum) {
      maxSum -= sum[v][i];
      v = up[v][i];
      cnt += 1 << i;
    }
  return cnt;
}
int dfs2(int v, int anc = 0) {
  int res = 0, tmp;
  for (int i = 0; i < graph[v].size(); i++)
    if (graph[v][i] != anc) {
      tmp = dfs2(graph[v][i], v);
      res = max(res, tmp);
    }
  if (res < 1) {
    ans++;
    return min(L, getMaxDist(v, S)) - 1;
  }
  return res - 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> L >> S;
  w.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    if (w[i] > S) {
      cout << "-1";
      return 0;
    }
  }
  if (n == 1) {
    cout << "1";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    up[i].resize(len + 1);
    sum[i].resize(len + 1);
  }
  for (int i = 1; i < n; i++) {
    cin >> l;
    l--;
    graph[i].push_back(l);
    graph[l].push_back(i);
  }
  dfs(0);
  dfs2(0);
  cout << ans;
  return 0;
}
