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;
inline long long getint() {
  long long ssum = 0, ff = 1;
  char ch;
  for (ch = getchar(); !isdigit(ch) && ch != '-'; ch = getchar())
    ;
  if (ch == '-') ff = -1, ch = getchar();
  for (; isdigit(ch); ch = getchar()) ssum = ssum * 10 + ch - '0';
  return ssum * ff;
}
const long long M = 2e5 + 5;
long long etot, head[M], go[M], nex[M];
long long s[M][21], f[M][21], n, m, L, S, mx, w[M], ans[M];
long long up[M];
inline void Addedge(long long u, long long v) {
  nex[++etot] = head[u];
  head[u] = etot;
  go[etot] = v;
}
void dfsbuild(long long u, long long fa) {
  f[u][0] = fa;
  s[u][0] = w[u];
  for (long long i = 1; i <= 20; i++) f[u][i] = f[f[u][i - 1]][i - 1];
  for (long long i = 1; i <= 20; i++)
    s[u][i] = s[u][i - 1] + s[f[u][i - 1]][i - 1];
  for (long long i = head[u]; i; i = nex[i]) {
    long long v = go[i];
    if (v == fa) continue;
    dfsbuild(v, u);
  }
}
void dfs(long long u, long long fa) {
  for (long long i = head[u]; i; i = nex[i]) {
    long long v = go[i];
    if (v == fa) continue;
    dfs(v, u);
    up[u] = max(up[u], up[v] - 1);
    ans[u] += ans[v];
  }
  if (up[u] == -1) {
    ans[u]++;
    long long len = 0, sum = w[u];
    long long now = f[u][0];
    for (long long i = 20; i >= 0; i--) {
      if (len + (1 << i) <= L && sum + s[now][i] <= S) {
        len += (1 << i);
        sum += s[now][i];
        now = f[now][i];
      }
    }
    up[u] = len;
  }
}
signed main() {
  cin >> n >> L >> S;
  L--;
  for (long long i = 1; i <= n; i++) {
    w[i] = getint();
    if (S < w[i]) {
      cout << -1;
      return 0;
    }
  }
  for (long long i = 2; i <= n; i++) {
    long long Fa = getint();
    Addedge(Fa, i);
  }
  dfsbuild(1, 0);
  memset(up, -1, sizeof(up));
  dfs(1, 0);
  cout << ans[1] << endl;
  return 0;
}
