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;
long long dis[101010];
int dep[101010];
int w[101010];
int h[101010], dp[101010];
int fa[101010][17];
int main() {
  int n, l, ans = 0;
  long long s;
  scanf("%d%d%lld", &n, &l, &s);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  dep[0] = 0;
  dep[1] = 1;
  memset(fa, 0, sizeof(fa));
  for (int i = 1; i <= n; i++) {
    if (i > 1) scanf("%d", &fa[i][0]);
    dep[i] = dep[fa[i][0]] + 1;
    dis[i] = dis[fa[i][0]] + w[i];
    for (int j = 1; j <= 16; j++) fa[i][j] = fa[fa[i][j - 1]][j - 1];
    h[i] = i;
    for (int j = 16; j >= 0; j--)
      if (dis[i] - dis[fa[h[i]][j]] <= s) h[i] = fa[h[i]][j];
    h[i] = min(l, dep[i] - dep[h[i]]);
  }
  memset(dp, 0, sizeof(dp));
  for (int i = n; i >= 1; i--) {
    if (dp[i] == 0) {
      dp[i] = h[i];
      ans++;
    }
    if (dp[i] == 0) {
      printf("-1");
      return 0;
    }
    dp[fa[i][0]] = max(dp[fa[i][0]], dp[i] - 1);
  }
  printf("%d", ans);
  return 0;
}
