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;
long long n, s, l, ans;
long long sppar[100010][30], up[100010], h[100010], chbeg[100010];
long long sum[100010], w[100010];
vector<vector<long long> > gr;
void precalc(long long v, long long par) {
  if (par == -1) {
    h[v] = 1;
    sum[v] = w[v];
  } else {
    h[v] = h[par] + 1;
    sum[v] = sum[par] + w[v];
  }
  sppar[v][0] = par;
  up[v] = v;
  for (long long i = 1; i < 20; ++i) {
    if (sppar[v][i - 1] == -1) {
      sppar[v][i] = -1;
      continue;
    }
    sppar[v][i] = sppar[sppar[v][i - 1]][i - 1];
  }
  long long ald = l - 1;
  for (long long i = 19; i >= 0; --i) {
    if (sppar[up[v]][i] == -1 or (1 << i) > ald) {
      continue;
    }
    if (sum[v] - sum[sppar[up[v]][i]] + w[sppar[up[v]][i]] <= s) {
      ald -= (1 << i);
      up[v] = sppar[up[v]][i];
    }
  }
  for (auto i : gr[v]) {
    precalc(i, v);
  }
}
long long dfsdp(long long v, long long par) {
  long long res = -1, tmpans = 0;
  for (auto i : gr[v]) {
    tmpans += dfsdp(i, v);
    if (h[chbeg[i]] > h[v]) {
      continue;
    }
    if (res < 0 or h[res] > h[chbeg[i]]) {
      res = chbeg[i];
    }
  }
  if (res < 0) {
    res = up[v];
    ++tmpans;
  }
  chbeg[v] = res;
  return tmpans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> l >> s;
  vector<long long> vi;
  gr.resize(n, vi);
  for (long long i = 0; i < n; ++i) {
    cin >> w[i];
    if (w[i] > s) {
      cout << -1;
      return 0;
    }
  }
  for (long long i = 1; i < n; ++i) {
    long long j;
    cin >> j;
    gr[j - 1].push_back(i);
  }
  precalc(0, -1);
  ans = dfsdp(0, -1);
  cout << ans << "\n";
}
