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 = 100005;
vector<int> G[MAXN];
int fa[25][MAXN];
int dep[MAXN];
int top[MAXN];
int W[MAXN];
long long sum[MAXN];
int son[MAXN];
long long N, L, S;
void initST() {
  fa[0][1] = -1;
  for (int k = 0; k < 20; k++)
    for (int i = 1; i <= N; i++)
      if (fa[k][i] < 0)
        fa[k + 1][i] = -1;
      else
        fa[k + 1][i] = fa[k][fa[k][i]];
}
void dfs1(int u, int pre) {
  sum[u] = sum[pre] + W[u];
  dep[u] = dep[pre] + 1;
  top[u] = u;
  int dis = L;
  for (int k = 20; k >= 0; k--) {
    int f = fa[k][top[u]];
    if ((1 << k) >= dis || f == -1) continue;
    if (sum[u] - sum[f] + W[f] > S) continue;
    dis -= (1 << k);
    top[u] = f;
  }
  for (int i = 0; i < G[u].size(); i++) dfs1(G[u][i], u);
}
int ans = 0;
void dfs2(int u) {
  int best = -1;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    dfs2(v);
    if (son[v] == v) continue;
    if (best == -1 || dep[best] > dep[son[v]]) best = son[v];
  }
  if (best == -1) {
    best = top[u];
    ans++;
  }
  son[u] = best;
}
int main() {
  scanf("%lld%lld%lld", &N, &L, &S);
  for (int i = 1; i <= N; i++) {
    scanf("%lld", &W[i]);
    if (W[i] > S) {
      cout << -1 << endl;
      return 0;
    }
  }
  int tmp;
  for (int i = 2; i <= N; i++) {
    scanf("%d", &tmp);
    fa[0][i] = tmp;
    G[tmp].push_back(i);
  }
  initST();
  dfs1(1, 0);
  dfs2(1);
  cout << ans << endl;
  return 0;
}
