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;
inline long long read() {
  long long x = 0, w = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * w;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long maxn = 1e5 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e9;
inline long long min(long long x, long long y) { return x < y ? x : y; }
inline long long max(long long x, long long y) { return x > y ? x : y; }
struct node {
  long long nxt, to;
} edge[maxn << 1];
long long n, l, s, tot, head[maxn];
long long dep[maxn], dep2[maxn], fa[maxn][25];
long long val[maxn], vis[maxn], len[maxn], ans[maxn], f[maxn];
inline void addedge(long long u, long long v) {
  edge[++tot] = (node){head[u], v}, head[u] = tot;
}
inline void dfs1(long long u, long long fath) {
  dep2[u] = dep2[fath] + val[u];
  fa[u][0] = fath, dep[u] = dep[fath] + 1;
  for (long long i = 1; i <= 20; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (long long i = head[u]; i; i = edge[i].nxt)
    if (edge[i].to != fath) dfs1(edge[i].to, u);
}
inline void dfs2(long long u) {
  f[u] = -1, ans[u] = 0;
  if (!head[u]) {
    f[u] = len[u] - 1, ans[u] = 1;
    return;
  }
  for (long long i = head[u]; i; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (v == fa[u][0]) continue;
    dfs2(v), ans[u] += ans[v], f[u] = max(f[u], f[v] - 1);
  }
  if (f[u] == -1) f[u] = len[u] - 1, ans[u]++;
}
signed main() {
  n = read(), l = read(), s = read();
  for (long long i = 1; i <= n; i++) {
    val[i] = read();
    if (val[i] > s) return puts("-1"), 0;
  }
  for (long long i = 2, fath; i <= n; i++) {
    fath = read();
    addedge(fath, i), addedge(i, fath);
  }
  dfs1(1, 0);
  for (long long i = 1, u; i <= n; i++) {
    u = i;
    for (long long j = 20; j >= 0; j--) {
      if (fa[u][j] != 0 && dep[i] - dep[fa[fa[u][j]][0]] <= l &&
          dep2[i] - dep2[fa[fa[u][j]][0]] <= s)
        u = fa[u][j];
    }
    len[i] = dep[i] - dep[u] + 1;
  }
  dfs2(1);
  printf("%lld\n", ans[1]);
  return 0;
}
