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 + 7;
int w[N];
long long best[N], lim[N], pw[N][20];
vector<int> g[N];
int n, l, len;
int up[N][20], flat[N], plen[N];
long long s, ans;
void dfs(int v, int p, int flor = 0) {
  up[v][0] = p;
  pw[v][0] = w[p];
  flat[v] = flor;
  for (int i = 1; i <= len; i++) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    pw[v][i] = pw[v][i - 1] + pw[up[v][i - 1]][i - 1];
  }
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    dfs(to, v, flor + 1);
  }
}
bool can_it(int a, int b, int i, long long path) {
  return (plen[a] + flat[a] - flat[b] <= l && (pw[a][i] + best[a] + path) <= s);
}
int bin_jump(int v) {
  int a = v;
  long long path = 0;
  for (int i = len; i >= 0; i--) {
    if (up[v][i] && can_it(a, up[v][i], i, path)) {
      v = up[v][i];
      path += pw[v][i];
    }
  }
  return flat[a] - flat[v];
}
void dfs1(int v, int p) {
  long long mxv = 0, mx = 0;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    dfs1(to, v);
    if (mx < lim[to]) {
      mx = lim[to];
      mxv = to;
    }
  }
  ans += (int(g[v].size()) - 2);
  ans += (v == 1);
  if (mxv == 0) ans++;
  best[v] = best[mxv] + w[v];
  plen[v] = plen[mxv] + 1;
  lim[v] = bin_jump(v);
}
int main() {
  cin >> n >> l >> s;
  len = log2(n) + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    if (w[i] > s) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    int v;
    scanf("%d", &v);
    g[i].push_back(v);
    g[v].push_back(i);
  }
  dfs(1, 0);
  dfs1(1, 0);
  cout << ans + 1 << endl;
}
