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 = 100005;
const int logN = 20;
vector<int> t[N];
int anc[N][logN], sum[N][logN], p[N], d[N], cnt[N], l;
long long dp[N][2], w[N], s;
bool valid = true;
void dfs0(int v, int h) {
  d[v] = h;
  int curr = anc[v][0] = p[v];
  sum[v][0] = w[0];
  for (int i = 1; i < logN; i++) {
    anc[v][i] = anc[curr][i - 1];
    sum[v][i] = sum[v][i - 1] + sum[curr][i - 1];
    curr = anc[curr][i - 1];
  }
  for (auto u : t[v]) dfs0(u, h + 1);
}
long long up(int v, long long tot) {
  if (tot < 0) return d[v] + 1;
  for (int i = logN - 1; i >= 0; i--)
    if (sum[v][i] <= tot) {
      tot -= sum[v][i];
      v = anc[v][i];
    }
  return d[v];
}
void dfs1(int v) {
  int ans = d[v], nxt = -1;
  for (auto u : t[v]) {
    dfs1(u);
    cnt[v] += cnt[u];
    int val = max(up(v, s - dp[u][0] - w[v]), dp[u][1]);
    if (val <= ans) ans = val, nxt = u;
  }
  if (w[v] > s) valid = false;
  if (nxt == -1)
    dp[v][0] = w[v], dp[v][1] = d[v] - l + 1, cnt[v]++;
  else
    dp[v][0] = dp[nxt][0] + w[v], dp[v][1] = dp[nxt][1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n >> l >> s;
  for (int i = 1; i <= n; i++) cin >> w[i];
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
    t[p[i]].push_back(i);
  }
  dfs0(1, 0);
  dfs1(1);
  cout << (valid ? cnt[1] : -1) << endl;
}
