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 = 100001;
const int M = N << 1;
const int Lg = 20;
int n, L;
long long S;
int cnt, head[N], Next[M], v[M];
int w[N];
int up[N];
int dep[N], f[N][Lg];
long long val[N][Lg];
int dp[N], mn[N];
void read(int &x) {
  char ch = getchar();
  x = 0;
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
}
void read(long long &x) {
  char ch = getchar();
  x = 0;
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
}
void add(int x, int y) {
  Next[++cnt] = head[x];
  head[x] = cnt;
  v[cnt] = y;
}
void dfs(int x, int fa) {
  dep[x] = dep[fa] + 1;
  f[x][0] = fa;
  val[x][0] = w[fa];
  for (int i = 1; i < Lg; i++) {
    f[x][i] = f[f[x][i - 1]][i - 1];
    val[x][i] = val[x][i - 1] + val[f[x][i - 1]][i - 1];
  }
  for (int i = head[x]; i; i = Next[i]) dfs(v[i], x);
}
void findup(int x) {
  if (w[x] > S) {
    printf("-1\n");
    exit(0);
  }
  int y = x;
  long long s = w[x];
  for (int i = Lg - 1; i >= 0; i--)
    if (s + val[y][i] <= S) s += val[y][i], y = f[y][i];
  if (y == 0) y = 1;
  up[x] = dep[x] - dep[y];
  up[x] = min(up[x], L);
}
void ddfs(int x, int fa) {
  for (int i = head[x]; i; i = Next[i]) {
    ddfs(v[i], x);
    mn[x] += mn[v[i]];
    dp[x] = max(dp[x], dp[v[i]]);
  }
  if (dp[x] > 0)
    dp[x]--;
  else
    mn[x]++, dp[x] = up[x];
}
int main() {
  read(n);
  read(L);
  read(S);
  L--;
  for (int i = 1; i <= n; i++) read(w[i]);
  int x;
  for (int i = 2; i <= n; i++) {
    read(x);
    add(x, i);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) findup(i);
  ddfs(1, 0);
  cout << mn[1] << endl;
  return 0;
}
