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 N = 1e5 + 5;
const int M = 20;
vector<int> adj[N];
int jump[N][M];
long long sum[N][M];
int depth[N];
long long query(int node, int l) {
  long long ret = 0;
  for (int i = 0; i < M; ++i) {
    if (l & (1 << i)) {
      ret += sum[node][i];
      node = jump[node][i];
    }
  }
  return ret;
}
int ret[N];
int len[N];
long long W[N];
void dfs(int node, int L, long long S) {
  int R = 1;
  bool ok = false;
  len[node] = 1;
  W[node] = sum[node][0];
  int canGo = 0;
  for (int i = 0; i < adj[node].size(); ++i) {
    int ch = adj[node][i];
    jump[ch][0] = node;
    depth[ch] = depth[node] + 1;
    for (int j = 1; j < M; ++j) {
      jump[ch][j] = jump[jump[ch][j - 1]][j - 1];
      sum[ch][j] = sum[ch][j - 1] + sum[jump[ch][j - 1]][j - 1];
    }
    dfs(ch, L, S);
    R += ret[ch];
    int rem = min(depth[node], L - len[ch]);
    if (rem == 0 || W[ch] + sum[node][0] > S) {
      continue;
    }
    int l = 1, r = rem, mid;
    while (l < r) {
      mid = (l + r + 1) / 2;
      if (query(node, mid) + W[ch] <= S) {
        l = mid;
      } else {
        r = mid - 1;
      }
    }
    if (!ok) {
      canGo = l;
      ok = true;
      --R;
      len[node] = len[ch] + 1;
      W[node] = W[ch] + sum[node][0];
    }
    if (ok && canGo < l) {
      canGo = l;
      len[node] = len[ch] + 1;
      W[node] = W[ch] + sum[node][0];
    }
  }
  ret[node] = R;
}
void solve() {
  int n, L;
  long long S;
  scanf("%d%d%lld", &n, &L, &S);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &sum[i][0]);
    if (sum[i][0] > S) {
      cout << -1 << endl;
      return;
    }
  }
  for (int i = 2; i <= n; ++i) {
    int p;
    scanf("%d", &p);
    adj[p].push_back(i);
  }
  depth[1] = 1;
  dfs(1, L, S);
  cout << ret[1] << endl;
}
int main() { solve(); }
