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 maxn = 1e5 + 10;
int n, L, f[maxn], d[maxn], nxt[maxn], g[maxn];
long long S, w[maxn], sum[maxn];
vector<int> E[maxn], stk;
void make(int x) {
  stk.push_back(x);
  for (int i = 0; i < E[x].size(); i++) {
    int y = E[x][i];
    sum[y] = sum[x] + w[y];
    d[y] = d[x] + 1;
    make(y);
  }
  int l = max(1, d[x] - L + 1), r = d[x], res;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (sum[x] - sum[stk[mid - 1]] <= S)
      res = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  nxt[x] = stk[res];
  stk.pop_back();
}
void DFS(int x) {
  int cur = 0;
  if (E[x].empty()) {
    f[x] = 1;
    g[x] = nxt[x];
    return;
  }
  for (int i = 0; i < E[x].size(); i++) {
    int y = E[x][i];
    DFS(y);
    if (cur == 0 || d[g[y]] < d[g[cur]]) cur = y;
    f[x] += f[y];
  }
  if (d[g[cur]] > d[x])
    g[x] = nxt[x], f[x]++;
  else
    g[x] = g[cur];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> L >> S;
  for (int i = 1; i <= n; i++) cin >> w[i];
  for (int i = 1; i <= n; i++)
    if (w[i] > S) return 0 & puts("-1");
  for (int i = 2, x; i <= n; i++) {
    cin >> x;
    E[x].push_back(i);
  }
  stk.push_back(0);
  d[1] = 1;
  sum[1] = w[1];
  make(1);
  g[0] = n;
  DFS(1);
  cout << f[1];
}
