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>
#pragma comment(linker, "/STACK:367772116")
using namespace std;
vector<long long> v;
int n, t;
long long s;
vector<vector<int>> e;
vector<int> pr;
vector<int> nabr;
long long curpr = 0;
struct str {
  int val = 0;
  str* l;
  str* r;
};
long long from = 0;
long long too = 2 << 60;
vector<vector<long long>> dusum;
vector<vector<int>> duvert;
vector<int> lvl;
void dfs1(int pos, int last, int curlvl) {
  int curup = 1;
  int curvert = pos;
  int num = 1;
  duvert[0][pos] = pos;
  dusum[0][pos] = v[pos];
  if (curlvl >= 1) {
    duvert[1][pos] = curvert = duvert[0][last];
    dusum[1][pos] = v[pos] + dusum[0][last];
    num = 2;
    curup = 4;
  }
  while (true) {
    if (curup > curlvl + 1) break;
    duvert[num][pos] = curvert = duvert[num - 1][curvert];
    dusum[num][pos] = dusum[num - 1][pos] + dusum[num - 1][curvert];
    num++;
    curup <<= 1;
  }
  for (auto to : e[pos]) {
    if (to == last) continue;
    dfs1(to, pos, curlvl + 1);
  }
}
int ans = 0;
int dfs2(int pos, int last, int curlvl) {
  int bestnum = -1;
  int id = -1;
  for (auto to : e[pos]) {
    if (to == last) continue;
    int cur = dfs2(to, pos, curlvl + 1);
    bestnum = max(cur, bestnum);
  }
  if (bestnum == -1) {
    ans++;
    long long ost = s;
    int l = 0;
    int r = min(t, curlvl + 1) + 1;
    while (r - l > 1) {
      int m = l + r >> 1;
      int m2 = m;
      long long cursum = 0;
      int curvert = pos;
      for (int i = 25; i >= 0; i--) {
        if (m & (1 << i)) {
          cursum += dusum[i][curvert];
          curvert = duvert[i][curvert];
        }
      }
      if (cursum > s)
        r = m;
      else
        l = m;
    }
    int num = l;
    bestnum = num - 1;
  }
  return bestnum - 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> t >> s;
  v.resize(n);
  e.resize(n);
  nabr.resize(n);
  for (size_t i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] > s) {
      cout << -1;
      return 0;
    }
  }
  lvl.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int a;
    cin >> a;
    a--;
    e[i + 1].push_back(a);
    e[a].push_back(i + 1);
  }
  dusum.resize(log2(s) + 2, vector<long long>(n, -1));
  duvert.resize(log2(s) + 2, vector<int>(n, -1));
  dfs1(0, -1, 0);
  dfs2(0, -1, 0);
  cout << ans;
}
