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;
int n, m, ans, h[200000], a[200000], d[200000], f[200000][18];
long long k, s[200000], w[200000][18];
int main() {
  scanf("%d%d%I64d", &n, &m, &k);
  f[1][0] = 1;
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &s[i]);
    if (s[i] > k) {
      printf("-1");
      return 0;
    }
  }
  for (int i = (2); i <= (n); i++) {
    scanf("%d", &f[i][0]);
    w[i][0] = s[f[i][0]];
    h[i] = h[f[i][0]] + 1;
  }
  for (int j = (1); j <= (17); j++)
    for (int i = (1); i <= (n); i++) {
      f[i][j] = f[f[i][j - 1]][j - 1];
      w[i][j] = w[i][j - 1] + w[f[i][j - 1]][j - 1];
    }
  for (int i = (1); i <= (n); i++) {
    int x = i, y = m - 1;
    long long z = k - s[i];
    for (int j = (17); j >= (0); j--)
      if ((y >= (1 << j)) && (z >= w[x][j])) {
        y -= 1 << j;
        z -= w[x][j];
        x = f[x][j];
      }
    d[i] = x;
  }
  for (int i = (n); i >= (1); i--) {
    if ((!a[i]) || (h[a[i]] > h[i])) a[i] = d[i], ans++;
    if ((i > 1) && ((!a[f[i][0]]) || (h[a[i]] < h[a[f[i][0]]])))
      a[f[i][0]] = a[i];
  }
  printf("%d", ans);
}
