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 Max = 1e5 + 3, Log = 20 + 1;
vector<int> adj[Max];
int memo[Max][Log], value[Max];
long long memoSum[Max][Log] = {0};
int depth[Max], dp[Max], sumDp[Max], highest[Max], path[Max];
void dfs(int node, int prev, int Depth = 0) {
  depth[node] = Depth;
  memo[node][0] = prev;
  memoSum[node][0] = value[node];
  for (int i = 1; i < Log; ++i) {
    memo[node][i] = memo[memo[node][i - 1]][i - 1];
    memoSum[node][i] = memoSum[node][i - 1] + memoSum[memo[node][i - 1]][i - 1];
  }
  for (int to : adj[node]) {
    if (to == prev) continue;
    dfs(to, node, Depth + 1);
  }
}
int solve(int node, int prev) {
  int ans = 0;
  int best = -1;
  for (int u : adj[node]) {
    if (u == prev) continue;
    ans += solve(u, node);
    if (path[u] == u) continue;
    if (best == -1 || depth[best] > depth[path[u]]) best = path[u];
  }
  if (best == -1) {
    best = highest[node];
    ++ans;
  }
  path[node] = best;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, l;
  long long s;
  cin >> n >> l >> s;
  for (int i = 1; i <= n; ++i) {
    dp[i] = n + 2;
    cin >> value[i];
    if (value[i] > s) {
      cout << "-1\n";
      return 0;
    }
  }
  int par;
  for (int i = 2; i <= n; ++i) {
    cin >> par;
    adj[par].push_back(i);
    adj[i].push_back(par);
  }
  dfs(1, 1);
  for (int i = 1; i <= n; ++i) {
    int node = i;
    long long sum = 0;
    int currDist = 0;
    for (int j = Log - 1; j >= 0; --j) {
      if (currDist + (1 << j) < l &&
          sum + memoSum[node][j] + value[memo[node][j]] <= s) {
        sum += memoSum[node][j];
        currDist += (1 << j);
        node = memo[node][j];
      }
    }
    highest[i] = node;
  }
  cout << solve(1, 1) << '\n';
  return 0;
}
