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;
template <typename T>
inline T rd() {
  T nmb = 0;
  int sgn = 0;
  char chr = getchar();
  while (!isdigit(chr)) {
    if (chr == '-') sgn = 1;
    chr = getchar();
  }
  while (isdigit(chr)) {
    nmb = (nmb << 3) + (nmb << 1) + chr - '0';
    chr = getchar();
  }
  return sgn ? -nmb : nmb;
}
template <typename T>
void wt(T nmb) {
  if (nmb > 9) wt(nmb / 10);
  putchar(nmb % 10 + '0');
}
template <typename T>
inline void cmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
inline void cmin(T &x, T y) {
  x = x < y ? x : y;
}
inline void proc_status() {
  ifstream t("/proc/self/status");
  cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>())
       << endl;
}
const int N = 1e5 + 10, K = 20;
int L;
int p[N];
long long S;
vector<int> g[N];
int dep[N], fa[N][K];
long long d[N];
void init(int x) {
  dep[x] = dep[fa[x][0]] + 1;
  for (int i = 1; i <= 16; ++i) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  d[x] = d[fa[x][0]] + p[x];
  for (int i = g[x].size() - 1; ~i; --i) {
    int k = g[x][i];
    init(k);
  }
}
inline int climb(int x, int p) {
  for (int i = 16; ~i; --i)
    if (p >= (1 << i)) x = fa[x][i], p -= (1 << i);
  return x;
}
int len[N];
long long sum[N];
int Ans;
void dfs(int x) {
  int mx = 0, pos = 0;
  for (int i = g[x].size() - 1; ~i; --i) {
    int k = g[x][i];
    dfs(k);
    int lft = 1, rgt = min(dep[x], L - len[k]), ans = 0;
    while (lft <= rgt) {
      int mid = (lft + rgt) >> 1;
      int t = climb(k, mid);
      if (sum[k] + d[x] - d[fa[t][0]] <= S)
        ans = mid, lft = mid + 1;
      else
        rgt = mid - 1;
    }
    if (mx < ans) mx = ans, pos = k;
  }
  if (pos)
    Ans += g[x].size() - 1, len[x] = len[pos] + 1, sum[x] = sum[pos] + p[x];
  else
    Ans += g[x].size(), len[x] = 1, sum[x] = p[x];
}
int main() {
  int n = rd<int>();
  L = rd<int>(), S = rd<long long>();
  for (int i = 1; i <= n; ++i) {
    p[i] = rd<int>();
    if (p[i] > S) return puts("-1"), 0;
  }
  for (int x = 2; x <= n; ++x) {
    int f = rd<int>();
    g[f].push_back(x), fa[x][0] = f;
  }
  init(1);
  dfs(1);
  ++Ans;
  printf("%d\n", Ans);
  return 0;
}
