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>
void read(T &x) {
  T f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ '0');
    s = getchar();
  }
  x *= f;
}
template <typename T>
void write(T x, char s = '\n') {
  if (!x) {
    putchar('0');
    putchar(s);
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  T t = 0, tmp[25] = {};
  while (x) tmp[t++] = x % 10, x /= 10;
  while (t-- > 0) putchar(tmp[t] + '0');
  putchar(s);
}
const int MAXN = 2e5 + 5;
int n, L, dep[MAXN], tag[MAXN], dp[MAXN], f[MAXN][22];
long long S, a[MAXN], Dis[MAXN];
int head[MAXN], to[MAXN << 1], nxt[MAXN << 1], cnt;
void add(int u, int v) {
  nxt[++cnt] = head[u];
  head[u] = cnt;
  to[cnt] = v;
}
void pre(int x, int fa) {
  dep[x] = dep[fa] + 1;
  Dis[x] = Dis[fa] + a[x];
  for (int i = 1; i <= 20; ++i) f[x][i] = f[f[x][i - 1]][i - 1];
  for (int i = head[x]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    f[v][0] = x;
    pre(v, x);
  }
}
void dfs(int x, int fa) {
  for (int i = head[x]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    dfs(v, x);
    dp[x] += dp[v];
    tag[x] += tag[v];
  }
  if (!tag[x]) {
    dp[x]++;
    int y = x;
    for (int i = 20; i >= 0; --i) {
      int p = f[y][i];
      if (dep[x] - dep[p] <= L && Dis[x] - Dis[p] <= S) y = f[y][i];
    }
    tag[x]++, tag[y]--;
  }
}
int main() {
  read(n), read(L), read(S);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    if (a[i] > S) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 2; i <= n; ++i) {
    int p;
    read(p);
    add(p, i);
  }
  pre(1, 0);
  dfs(1, 0);
  write(dp[1]);
  return 0;
}
