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, l, s;
long long w[100005];
long long res;
vector<long long> niz[100005], brisi[100005];
long long ulaz[100005], izlaz[100005], dubina[100005], vreme;
void dfsU(long long t, long long h) {
  vreme++;
  dubina[t] = h;
  ulaz[t] = vreme;
  for (auto x : niz[t]) dfsU(x, h + 1);
  izlaz[t] = vreme;
}
long long cale[100005][21], zbir[100005][21];
long long nadji(long long t) {
  long long mdub = max(0ll, dubina[t] - l), tz = s;
  if (!mdub && zbir[t][20] <= s) return 0;
  for (long long j = 20; j >= 0; j--) {
    if (cale[t][j] == 0) continue;
    if (dubina[cale[t][j]] < mdub || zbir[t][j] > tz) continue;
    tz -= zbir[t][j];
    t = cale[t][j];
  }
  return t;
}
long long bit[100005];
void dodaj(long long p, long long x) {
  for (; p < 100005; p += -p & (p)) bit[p] += x;
}
long long dobij(long long p) {
  long long z = 0;
  for (; p > 0; p -= (-p) & p) z += bit[p];
  return z;
}
void dfs(long long t) {
  for (auto x : niz[t]) dfs(x);
  for (auto x : brisi[t]) dodaj(x, -1);
  if (dobij(izlaz[t]) - dobij(ulaz[t])) return;
  res++;
  dodaj(ulaz[t], 1);
  brisi[nadji(t)].push_back(ulaz[t]);
}
int main() {
  scanf("%I64d %I64d %I64d", &n, &l, &s);
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", w + i);
    if (w[i] > s) {
      printf("-1");
      return 0;
    }
  }
  for (long long i = 2; i <= n; i++) {
    long long t;
    scanf("%I64d", &t);
    niz[t].push_back(i);
    cale[i][0] = t;
    zbir[i][0] = w[i];
  }
  zbir[1][0] = w[1];
  for (long long j = 1; j < 21; j++) {
    for (long long i = 1; i <= n; i++) {
      cale[i][j] = cale[cale[i][j - 1]][j - 1];
      zbir[i][j] = zbir[cale[i][j - 1]][j - 1] + zbir[i][j - 1];
    }
  }
  dfsU(1, 1);
  dfs(1);
  printf("%I64d", res);
  return 0;
}
