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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.000000000000000000001;
using namespace std;
int I, T[100005], A[100005][18], L[100005], R[100005];
pair<int, int> DP[100005];
long long int S, W[100005][18];
vector<int> P[100005];
void dfs(int n, int p) {
  for (auto v : P[n]) {
    if (v != p) {
      L[v] = L[n] + 1;
      A[v][0] = n;
      W[v][0] = T[n];
      dfs(v, n);
    }
  }
}
void dfs2(int n, int p) {
  int s = 0, l = inf;
  bool czy = 0;
  for (auto v : P[n]) {
    if (v != p) {
      dfs2(v, n);
      s += DP[v].first;
      if (DP[v].second <= L[n]) {
        czy = 1;
        l = min(l, DP[v].second);
      }
    }
  }
  if (!czy) s++, l = R[n];
  DP[n] = make_pair(s, l);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n;
  cin >> n >> I >> S;
  for (int i = 1; i <= n; i++) {
    cin >> T[i];
    if (T[i] > S) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    int a;
    cin >> a;
    P[a].push_back(i);
    P[i].push_back(a);
  }
  A[1][0] = 1;
  dfs(1, 0);
  for (int i = 1; i <= 17; i++) {
    for (int j = 1; j <= n; j++) {
      A[j][i] = A[A[j][i - 1]][i - 1];
      W[j][i] = W[j][i - 1] + W[A[j][i - 1]][i - 1];
    }
  }
  for (int i = 1; i <= n; i++) {
    int l = 1;
    long long int s = T[i];
    int a = i;
    for (int j = 17; j >= 0; j--) {
      if ((1 << j) + l <= I && s + W[a][j] <= S) {
        l += (1 << j);
        s += W[a][j];
        a = A[a][j];
      }
    }
    R[i] = L[a];
  }
  dfs2(1, 0);
  cout << DP[1].first;
  return 0;
}
