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 MAXN = 1e5;
const int MAXL = 1e5;
struct Answer {
  int paths, length;
  long long sum;
};
long long acc[MAXN + 5];
vector<int> G[MAXN + 5];
int W[MAXN + 5];
int n, L;
long long S;
int getUp(int u, int dep, const Answer& child) {
  int s = 0, e = min(dep, L - child.length);
  int ret = 0;
  while (s <= e) {
    int mid = s + e >> 1;
    if (acc[dep] - acc[dep - mid] + child.sum <= S) {
      s = mid + 1;
      ret = mid;
    } else
      e = mid - 1;
  }
  return ret;
}
Answer solve(int u = 0, int dep = 1) {
  acc[dep] = acc[dep - 1] + W[u];
  Answer ret = {0, 0, 0};
  int bestUp = 0;
  for (int v : G[u]) {
    Answer child = solve(v, dep + 1);
    int up = getUp(u, dep, child);
    ret.paths += child.paths;
    if (up > bestUp) {
      ret.length = child.length;
      ret.sum = child.sum;
      bestUp = up;
    }
  }
  ret.length++;
  ret.sum += W[u];
  ret.paths += bestUp == 0;
  return ret;
}
int main() {
  int c, c2;
  scanf("%d%d%I64d", &n, &L, &S);
  for (c = 0; c < n; c++) {
    scanf("%d", &W[c]);
    if (W[c] > S) {
      printf("-1\n");
      return 0;
    }
  }
  for (c = 1; c < n; c++) {
    int p;
    scanf("%d", &p);
    G[p - 1].push_back(c);
  }
  acc[0] = 0;
  printf("%d\n", solve().paths);
  return 0;
}
