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;
int head[N], Next[N], ver[N], tot;
inline void add(int u, int v) {
  ver[++tot] = v, Next[tot] = head[u], head[u] = tot;
}
int pa[N][25], top[N], son[N], dep[N], ans, n, L;
long long sum[N], w[N], S;
inline void init(int u) {
  for (int i = 1; i <= 20; ++i) pa[u][i] = pa[pa[u][i - 1]][i - 1];
  int dis = L, tmp = u;
  for (int i = 20; i >= 0; --i) {
    int fa = pa[tmp][i];
    if (!fa || (1 << i) >= dis) continue;
    if (sum[u] - sum[fa] + w[fa] > S) continue;
    dis -= (1 << i), top[u] = fa, tmp = fa;
  }
}
void dfs1(int u, int fa) {
  sum[u] = sum[fa] + w[u], top[u] = u, dep[u] = dep[fa] + 1, init(u);
  for (int i = head[u]; i; i = Next[i]) dfs1(ver[i], u);
}
void dfs2(int u) {
  int mx = -1;
  for (int i = head[u]; i; i = Next[i]) {
    int v = ver[i];
    dfs2(v);
    if (son[v] == v) continue;
    if (mx == -1 || dep[mx] > dep[son[v]]) mx = son[v];
  }
  if (mx == -1) ++ans, mx = top[u];
  son[u] = mx;
}
int main() {
  scanf("%d%d%lld", &n, &L, &S);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &w[i]);
    if (w[i] > S) return puts("-1"), 0;
  }
  for (int i = 2, f; i <= n; ++i) scanf("%d", &f), add(f, i), pa[i][0] = f;
  dfs1(1, 0), dfs2(1);
  printf("%d\n", ans);
  return 0;
}
