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>
template <typename T>
inline void read(T &x) {
  x = 0;
  register char ch;
  register bool fl = 0;
  while (ch = getchar(), ch < 48 || 57 < ch) fl ^= ch == '-';
  x = (ch & 15);
  while (ch = getchar(), 47 < ch && ch < 58)
    x = (x << 1) + (x << 3) + (ch & 15);
  if (fl) x = -x;
}
template <typename T>
inline void readc(T &x) {
  while (x = getchar(), !islower(x) && !isupper(x))
    ;
}
template <typename T>
inline void print(T x, char c = ' ') {
  static int buf[40];
  if (x == 0) {
    putchar('0');
    putchar(c);
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  for (buf[0] = 0; x; x /= 10) buf[++buf[0]] = x % 10 + 48;
  while (buf[0]) putchar((char)buf[buf[0]--]);
  putchar(c);
}
const int maxn = 100010;
int n, m, x, ans;
int c[maxn], f[maxn][20], cnt[maxn], dp[maxn];
long long s, a[maxn], g[maxn][20], sum[maxn];
int tot = 2, hed[maxn], to[maxn << 1], nxt[maxn << 1];
inline void add_edge(int u, int v) {
  nxt[tot] = hed[u], to[tot] = v, hed[u] = tot++;
  nxt[tot] = hed[v], to[tot] = u, hed[v] = tot++;
}
void dfs1(int u) {
  for (int i = hed[u], v = to[i]; i; i = nxt[i], v = to[i])
    if (v ^ f[u][0]) {
      f[v][0] = u;
      g[v][0] = a[u];
      dfs1(v);
    }
}
int dfs2(int u) {
  int max = 0;
  for (int i = hed[u], v = to[i]; i; i = nxt[i], v = to[i])
    if (v ^ f[u][0]) {
      max = std::max(max, dfs2(v));
    }
  if (!max) {
    ans++;
    max = dp[u];
  }
  return max - 1;
}
int main() {
  read(n), read(m), 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++) {
    read(x);
    add_edge(x, i);
  }
  dfs1(1);
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++) {
      f[j][i] = f[f[j][i - 1]][i - 1];
      g[j][i] = g[f[j][i - 1]][i - 1] + g[j][i - 1];
    }
  for (int k = 1, u = k; k <= n; k++, u = k) {
    int stp = 1, sum = a[u];
    for (int i = 19; i >= 0; i--)
      if ((stp + (1 << i)) <= m && (sum + g[u][i]) <= s && f[u][i]) {
        stp += 1 << i;
        sum += g[u][i];
        u = f[u][i];
      }
    dp[k] = stp;
  }
  dfs2(1);
  print(ans, '\n');
  return 0;
}
