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 + 100;
vector<int> G[maxn];
int n, L, par[maxn], lg[maxn];
int w[maxn], f[maxn][22], dep[maxn];
long long cost[maxn][22];
long long S;
bool vis[maxn];
int Find(int x) {
  if (x == par[x]) return x;
  return par[x] = Find(par[x]);
}
priority_queue<pair<int, int> > que;
void dfs(int v, int fa) {
  dep[v] = dep[fa] + 1;
  que.push(pair<int, int>(dep[v], v));
  f[v][0] = fa;
  cost[v][0] = w[v];
  for (int i = 1; (1 << i) <= dep[v]; i++) {
    f[v][i] = f[f[v][i - 1]][i - 1];
    cost[v][i] = cost[v][i - 1] + cost[f[v][i - 1]][i - 1];
  }
  for (int i = 0; i < G[v].size(); i++) dfs(G[v][i], v);
}
int work() {
  int ans = 0;
  while (!que.empty()) {
    int v = que.top().second;
    que.pop();
    if (vis[v]) continue;
    int u = v;
    ans++;
    long long ss = S;
    int lll = L;
    for (int i = lg[dep[v]]; i >= 0; i--) {
      if (cost[u][i] && cost[u][i] <= ss && (1 << i) <= lll) {
        ss -= cost[u][i];
        lll -= (1 << i);
        u = f[u][i];
      }
    }
    while (dep[v] > dep[u]) {
      vis[v] = 1;
      par[v] = f[v][0];
      v = Find(v);
    }
  }
  return ans;
}
int main() {
  for (int i = 0; i < maxn; i++) par[i] = i;
  for (int i = 2; i < maxn; i++) lg[i] = lg[i / 2] + 1;
  int ma = 0;
  scanf("%d%d%I64d", &n, &L, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    ma = max(ma, w[i]);
  }
  if (ma > S) {
    printf("-1\n");
    return 0;
  }
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    G[x].push_back(i);
  }
  dfs(1, 0);
  printf("%d\n", work());
  return 0;
}
