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 long long N = 2e5 + 100;
const long long M = 21;
const long long mod = 1e9 + 7;
const long long MOD = 998244353;
const long double eps = 0.000000001;
const long long P = 1336;
const long long inf = 1e18;
const long long CONST = 5001;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
vector<long long> g[N];
long long a[N];
long long n, L, second;
long long up[N][M], h[N], sum[N];
pair<long long, long long> b[N];
void DFS(long long v, long long pr) {
  up[v][0] = pr;
  if (v == pr) up[v][0] = -1;
  h[v] = h[pr] + 1;
  sum[v] = sum[pr] + a[v];
  for (long long i = 1; i < M; i++) {
    if (h[v] < (1 << i) || up[v][i - 1] == -1) {
      up[v][i] = -1;
      continue;
    }
    up[v][i] = up[up[v][i - 1]][i - 1];
  }
  for (auto to : g[v]) {
    if (to != pr) DFS(to, v);
  }
}
long long f(long long v) {
  long long s = b[v].first, l = b[v].second;
  long long ans = 0;
  for (long long i = M - 1; i >= 0; i--) {
    long long x = up[v][i];
    if (x == -1 || x == v) continue;
    long long adds = sum[v] - sum[x] - a[v] + a[x];
    long long addl = h[v] - h[x];
    if (s + adds <= second && l + addl <= L) {
      ans += (1 << i);
      s += adds;
      l += addl;
      v = up[v][i];
    }
  }
  return ans;
}
void dfs(long long v, long long pr) {
  long long ma = 0;
  long long nom = -1;
  for (auto to : g[v]) {
    if (to == pr) continue;
    dfs(to, v);
    long long c = f(to);
    if (c > ma && b[to].first + a[v] <= second && b[to].second + 1 <= L) {
      ma = c;
      nom = to;
    }
  }
  if (nom == -1) {
    if (a[v] <= second && L) {
      b[v] = {a[v], 1};
      return;
    }
    cout << -1;
    exit(0);
    return;
  }
  b[v].first = b[nom].first + a[v];
  b[v].second = b[nom].second + 1;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  cin >> n >> L >> second;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i < n; i++) {
    long long x;
    cin >> x;
    x--;
    g[x].push_back(i);
    g[i].push_back(x);
  }
  DFS(0, 0);
  dfs(0, 0);
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    if (b[i].second == 1) ans++;
  }
  cout << ans;
}
