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;
template <typename T>
inline T read(T& x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
const int N = 1e5 + 10;
int n, i, j, L, v[N], dep[N], in[N], p[N], parent[N][18];
long long S, w[N], ss[N][18];
bool vis[N];
vector<int> G[N];
priority_queue<pair<int, int> > Q;
void dfs(int u, int f) {
  dep[u] = dep[f] + 1;
  for (int i = 0; i < (int)G[u].size(); ++i) {
    int v = G[u][i];
    if (v == f) continue;
    dfs(v, u);
  }
}
void preprocess() {
  for (i = 1; i <= n; ++i) {
    parent[i][0] = p[i];
    ss[i][0] = w[p[i]];
  }
  for (j = 1; j < 18; ++j) {
    for (i = 1; i <= n && i + (1 << j) <= n; ++i) {
      parent[i][j] = parent[parent[i][j - 1]][j - 1];
      ss[i][j] = ss[i][j - 1] + ss[parent[i][j - 1]][j - 1];
    }
  }
}
int main() {
  read(n), read(L), read(S);
  for (i = 1; i <= n; ++i) {
    read(w[i]);
    if (w[i] > S) return puts("-1"), 0;
  }
  for (i = 2; i <= n; ++i) {
    read(p[i]);
    G[p[i]].emplace_back(i);
    in[p[i]]++;
  }
  dfs(1, 0);
  preprocess();
  for (i = 1; i <= n; ++i) {
    long long limit = L - 1, slimit = S - w[i];
    int ins = 17, x = i;
    while (ins >= 0) {
      if (parent[x][ins] == 0)
        ins--;
      else if ((1 << ins) > limit)
        ins--;
      else if (ss[x][ins] <= slimit) {
        slimit -= ss[x][ins];
        limit -= (1 << ins);
        x = parent[x][ins];
      } else
        ins--;
    }
    v[i] = x;
  }
  for (i = 1; i <= n; ++i)
    if (in[i] == 0) Q.push(make_pair(dep[i], i));
  int ans = 0;
  while (!Q.empty()) {
    int u = Q.top().second;
    Q.pop();
    if (vis[u]) continue;
    int bel = v[u];
    int mid = p[u], pre = u;
    while (dep[mid] > dep[bel]) {
      vis[pre] = 1;
      p[pre] = bel;
      pre = mid;
      mid = p[mid];
    }
    vis[pre] = vis[bel] = 1;
    if (pre != bel) p[pre] = bel;
    ans++;
    if (p[bel] > 0 && !vis[p[bel]]) {
      Q.push(make_pair(dep[p[bel]], p[bel]));
    }
  }
  printf("%d\n", ans);
  return 0;
}
