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 _ = 1e5 + 5;
inline long long read() {
  char ch = '!';
  long long z = 1, num = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') z = -1, ch = getchar();
  while (ch <= '9' && ch >= '0')
    num = (num << 3) + (num << 1) + ch - '0', ch = getchar();
  return z * num;
}
int n, L, w[_];
struct ed {
  int to, next;
} e[_];
int cnt, head[_];
void link(int u, int v) {
  e[++cnt] = (ed){v, head[u]};
  head[u] = cnt;
}
int f[_][20];
long long g[_][20], s;
void dfs(int u) {
  for (int i = 1; i <= 17; ++i)
    f[u][i] = f[f[u][i - 1]][i - 1],
    g[u][i] = g[f[u][i - 1]][i - 1] + g[u][i - 1];
  for (int i = head[u]; i; i = e[i].next) g[e[i].to][0] = w[u], dfs(e[i].to);
}
int dp[_], Ans;
int Get(int u) {
  int Mx = 0;
  for (int i = head[u]; i; i = e[i].next) Mx = max(Mx, Get(e[i].to));
  if (!Mx) Ans++, Mx = dp[u];
  return Mx - 1;
}
int main() {
  n = read(), L = read(), s = read();
  for (int i = 1; i <= n; ++i) w[i] = read();
  for (int i = 1; i <= n; ++i)
    if (w[i] > s) return puts("-1"), 0;
  for (int i = 2; i <= n; ++i) f[i][0] = read(), link(f[i][0], i);
  dfs(1);
  for (int i = 1; i <= n; ++i) {
    int u = i, now = 1;
    long long res = w[i];
    for (int j = 17; ~j; --j)
      if (f[u][j] && now + (1 << j) <= L && res + g[u][j] <= s) {
        now += (1 << j);
        res += g[u][j];
        u = f[u][j];
      }
    dp[i] = now;
  }
  Get(1);
  printf("%d\n", Ans);
  return 0;
}
