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;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
}
using namespace std;
int n, mxl;
long long S;
long long w[101000];
int fa[101000];
struct edge {
  int nxt, to;
} e[101000 << 1];
int head[101000], ecnt;
inline void addedge(int from, int to) {
  e[++ecnt] = (edge){head[from], to};
  head[from] = ecnt;
}
int stk[101000], stop;
int dep[101000];
long long sum[101000];
int L[101000];
long long s[101000];
int ans;
void dfs(int cur) {
  stk[++stop] = cur;
  sum[cur] = sum[fa[cur]] + w[cur];
  dep[cur] = dep[fa[cur]] + 1;
  int nw = stop;
  int mn = n + 1;
  for (int i = head[cur]; i; i = e[i].nxt) {
    int to = e[i].to;
    dfs(to);
    int l = 1, r = min(nw, L[to]), res = n + 1;
    while (l <= r) {
      int mid = (l + r) >> 1;
      int tmp = stk[nw - mid];
      if (sum[cur] - sum[tmp] <= s[to])
        l = mid + 1, res = nw - mid;
      else
        r = mid - 1;
    }
    if (res < mn) mn = res, L[cur] = L[to] - 1, s[cur] = s[to] - w[cur];
  }
  if (mn == n + 1) L[cur] = mxl - 1, s[cur] = S - w[cur], ++ans;
  --stop;
}
int main() {
  read(n), read(mxl), read(S);
  for (int i = 1; i <= n; ++i) {
    read(w[i]);
    if (w[i] > S) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 2; i <= n; ++i) read(fa[i]), addedge(fa[i], i);
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
