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 N = 1e5 + 9;
long long n, x, y, l, r, mid, p, second, L, w[N], dp[N][3], masir[N];
vector<long long> g[N];
long long binary_search(long long t, long long v, long long x) {
  l = max(1ll * -1, x - t - 1), r = x;
  while (l + 1 < r) {
    mid = (l + r) / 2;
    if (masir[x - 1] - masir[mid - 1] <= v)
      r = mid;
    else
      l = mid;
  }
  return r;
}
void dfs(long long x, long long dist) {
  if (!g[x].size()) {
    dp[x][0] = 1, dp[x][1] = 1, dp[x][2] = w[x];
    return;
  }
  long long mn = 3e5, an = -1;
  masir[dist] = w[x];
  if (dist) masir[dist] += masir[dist - 1];
  for (long long i = 0; i < g[x].size(); i++) {
    dfs(g[x][i], dist + 1);
    dp[x][0] += dp[g[x][i]][0];
    p = binary_search(L - dp[g[x][i]][1], second - dp[g[x][i]][2], dist + 1);
    if (p < mn) mn = p, an = g[x][i];
  }
  if (mn == dist + 1)
    dp[x][0]++, dp[x][1] = 1, dp[x][2] = w[x];
  else
    dp[x][1] = dp[an][1] + 1, dp[x][2] = dp[an][2] + w[x];
}
int main() {
  cin >> n >> L >> second;
  for (long long i = 1; i < n + 1; i++) {
    scanf("%I64d", &w[i]);
    ;
    if (w[i] > second) return cout << -1, 0;
  }
  for (long long i = 2; i < n + 1; i++) {
    scanf("%I64d", &x);
    ;
    g[x].push_back(i);
  }
  dfs(1, 0);
  cout << dp[1][0];
}
