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 long long maxn = 100100;
long long head[maxn], nxt[maxn << 1], ver[maxn << 1], tot;
inline void addedge(long long a, long long b) {
  nxt[++tot] = head[a];
  ver[tot] = b;
  head[a] = tot;
  nxt[++tot] = head[b];
  ver[tot] = a;
  head[b] = tot;
}
long long n, w[maxn], L, S;
inline void init() {
  scanf("%lld%lld%lld", &n, &L, &S);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", w + i);
    if (w[i] > S) {
      puts("-1");
      exit(0);
    }
  }
  for (long long i = 2, x; i <= n; i++) scanf("%lld", &x), addedge(x, i);
}
long long fa[maxn][20], sum[maxn][20];
inline void dfs(long long x, long long fat) {
  fa[x][0] = fat;
  sum[x][0] = w[fat];
  for (long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == fat) continue;
    dfs(y, x);
  }
}
long long upv[maxn], f[maxn];
inline long long treedp(long long x, long long fat) {
  long long ans = 0;
  for (long long i = head[x]; i; i = nxt[i]) {
    long long y = ver[i];
    if (y == fat) continue;
    ans += treedp(y, x);
    f[x] = max(f[x], f[y] - 1);
  }
  if (f[x] == 0) ans++, f[x] = upv[x];
  return ans;
}
signed main() {
  init();
  dfs(1, 0);
  for (long long j = 1; j < 20; j++)
    for (long long i = 1; i <= n; i++) {
      fa[i][j] = fa[fa[i][j - 1]][j - 1];
      sum[i][j] = sum[i][j - 1] + sum[fa[i][j - 1]][j - 1];
    }
  for (long long i = 1; i <= n; i++) {
    long long val = S - w[i], resth = L - 1, now = i;
    for (long long j = 19; j >= 0; j--)
      if (fa[now][j] && resth - (1 << j) >= 0 && val - sum[now][j] >= 0) {
        val -= sum[now][j];
        resth -= 1 << j;
        now = fa[now][j];
      }
    upv[i] = L - resth;
  }
  printf("%lld\n", treedp(1, 0));
  return 0;
}
