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 INF = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5 + 10;
int n, L;
long long S;
int p[N];
vector<int> g[N];
long long w[N];
int dep[N];
const int LOGN = 20;
long long sum[LOGN][N];
int anc[LOGN][N];
void dfs(int u) {
  anc[0][u] = p[u];
  sum[0][u] = w[p[u]] + w[u];
  for (int k = 1; k < LOGN; ++k) {
    if ((1 << k) > dep[u]) break;
    anc[k][u] = anc[k - 1][anc[k - 1][u]];
    sum[k][u] = sum[k - 1][anc[k - 1][u]] + sum[k - 1][u] - w[anc[k - 1][u]];
  }
  for (auto v : g[u]) {
    dep[v] = dep[u] + 1;
    dfs(v);
  }
}
void init() {
  memset(sum, 0x3f, sizeof(sum));
  w[0] = INF64;
  dfs(1);
}
bool vis[N];
int ans;
int dfs2(int u) {
  int mn = INF;
  for (auto v : g[u]) {
    mn = min(mn, dfs2(v));
  }
  if (mn <= dep[u]) return mn;
  long long rem = S - w[u];
  int v = u;
  int l = L - 1;
  for (int k = LOGN - 1; k >= 0; --k) {
    if ((1 << k) > l || sum[k][v] - w[v] > rem) continue;
    rem -= sum[k][v] - w[v];
    l -= (1 << k);
    v = anc[k][v];
  }
  ++ans;
  do {
  } while (0);
  return dep[v];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> L >> S;
  for (int i = 1; i <= n; ++i) {
    cin >> w[i];
    if (w[i] > S) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 2; i <= n; ++i) {
    cin >> p[i];
    g[p[i]].push_back(i);
  }
  init();
  do {
  } while (0);
  do {
  } while (0);
  do {
  } while (0);
  do {
  } while (0);
  do {
  } while (0);
  do {
  } while (0);
  dfs2(1);
  cout << ans << '\n';
  return 0;
}
