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;
int n, l;
int64_t s;
vector<int> cost;
vector<vector<int>> ad;
int len;
vector<int64_t> ps;
vector<int> mx, id, dp;
void dfs(int node) {
  len++;
  ps[len] = ps[len - 1] + cost[node];
  int lo = max(0, len - l), hi = len + 1;
  while (hi - lo > 1) {
    int md = (lo + hi) / 2;
    if (ps[len] - ps[md - 1] > s)
      lo = md;
    else
      hi = md;
  }
  mx[node] = hi;
  for (int son : ad[node]) dfs(son);
  len--;
}
void dfs(int node, int depth) {
  int max = depth + 1, sum = 0;
  for (int son : ad[node]) {
    dfs(son, depth + 1);
    max = min(max, mx[id[son]]);
    sum += dp[son];
  }
  if (max == depth + 1) {
    id[node] = node;
    dp[node] = sum + 1;
    return;
  }
  for (int son : ad[node])
    if (mx[id[son]] == max) {
      id[node] = id[son];
      dp[node] = sum;
      return;
    }
}
int main() {
  cin >> n >> l >> s;
  ad.resize(n + 1);
  cost.resize(n + 1);
  ps.resize(n + 1);
  mx.resize(n + 1);
  id.resize(n + 1);
  dp.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> cost[i];
    if (cost[i] > s) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    ad[x].push_back(i);
  }
  dfs(1);
  dfs(1, 1);
  cout << dp[1] << '\n';
  return 0;
}
