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;
struct Edge {
  long long to, next;
} edge[100010 << 1];
long long num, head[100010], sum[100010], d[100010], fa[100010][25],
    top[100010], up[100010], ans, n, L, S, w[100010];
inline long long read() {
  long long s = 0, w = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') w = -1;
  for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
  return s * w;
}
void addedge(long long x, long long y) {
  edge[++num] = (Edge){y, head[x]};
  head[x] = num;
}
void dfs(long long u, long long pre) {
  sum[u] = sum[pre] + w[u], d[u] = d[pre] + 1;
  for (long long i = 0; fa[u][i]; ++i) fa[u][i + 1] = fa[fa[u][i]][i];
  top[u] = u;
  long long suml = L - 1;
  for (long long i = 20; i >= 0; --i) {
    long long tmp = top[u];
    if ((1 << i) > suml || !fa[tmp][i] ||
        sum[u] - sum[fa[tmp][i]] + w[fa[tmp][i]] > S)
      continue;
    suml -= (1 << i), top[u] = fa[tmp][i];
  }
  for (long long i = head[u]; i; i = edge[i].next) dfs(edge[i].to, u);
}
void dfs(long long u) {
  long long tmp = -1;
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    dfs(v);
    if (up[v] != v && (tmp == -1 || d[up[v]] < d[tmp])) tmp = up[v];
  }
  if (tmp == -1)
    ++ans, up[u] = top[u];
  else
    up[u] = tmp;
}
signed main() {
  n = read(), L = read(), S = read();
  for (long long i = 1; i <= n; ++i) {
    w[i] = read();
    if (w[i] > S) return puts("-1"), 0;
  }
  for (long long i = 2; i <= n; ++i) {
    fa[i][0] = read();
    addedge(fa[i][0], i);
  }
  dfs(1, 0);
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
