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 inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const double eps = 1e-8;
struct edge {
  int u, v, ne;
} e[maxn];
int n, al, ans;
int head[maxn], len;
long long s, w[maxn], dis[maxn], val[maxn];
int num[maxn];
int anc[maxn][21], f[maxn], lg[maxn], deep[maxn];
void add(int u, int v) {
  e[len].u = u;
  e[len].v = v;
  e[len].ne = head[u];
  head[u] = len++;
}
int get_fa(int x, int y) {
  for (int i = 19; i >= 0; i--)
    if (deep[x] - (1 << i) >= y) x = anc[x][i];
  return x;
}
void dfs(int x, int p) {
  anc[x][0] = p;
  f[x] = p;
  dis[x] = dis[p] + w[x];
  deep[x] = deep[p] + 1;
  for (int i = 1; i < 20; i++) anc[x][i] = anc[anc[x][i - 1]][i - 1];
  int maxh = 0, tag = -1;
  for (int i = head[x]; ~i; i = e[i].ne) {
    int v = e[i].v;
    if (v == p) continue;
    dfs(v, x);
    int l = 1, r = min(deep[x], al - num[v]);
    while (l <= r) {
      int mid = (l + r) >> 1;
      int fa = get_fa(x, deep[x] - mid);
      long long tmp = dis[x] - dis[f[fa]];
      if (tmp > s - val[v])
        r = mid - 1;
      else
        l = mid + 1;
    }
    ans++;
    if (r > maxh) {
      maxh = r;
      tag = v;
    }
  }
  if (tag == -1) {
    num[x] = 1;
    val[x] = w[x];
  } else {
    ans--;
    num[x] = num[tag] + 1;
    val[x] = val[tag] + w[x];
  }
  return;
}
void init() {
  lg[1] = 0;
  for (int i = 2; i <= 200000; i++) lg[i] = lg[i - 1] + (i & (i - 1)) == 0;
  return;
}
int main() {
  init();
  memset(head, -1, sizeof(head));
  cin >> n >> al >> s;
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    if (w[i] > s) flag = 0;
  }
  for (int i = 2, x; i <= n; i++) {
    scanf("%d", &x);
    add(x, i);
  }
  if (!flag) {
    cout << -1 << endl;
    return 0;
  }
  dfs(1, 0);
  ans++;
  cout << ans << endl;
  return 0;
}
