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;
long long s[200009], val[200009];
vector<int> son[200009];
int par[200009][20], lvl[200009];
long long remainingL[200009], remainingS[200009];
long long acum[200009][20];
long long S, L;
int ans = 0;
void dfs(int u) {
  for (int i = 0; i < int(son[u].size()); i++) {
    int v = son[u][i];
    dfs(v);
  }
  ans++;
  remainingS[u] = S - val[u];
  remainingL[u] = L - 1;
  int bestSon = -1, maxUp = -1;
  for (int i = 0; i < int(son[u].size()); i++) {
    int v = son[u][i];
    if (remainingS[v] < val[u]) continue;
    if (remainingL[v] == 0) continue;
    int k = v;
    long long sum = remainingS[v];
    for (int j = 20 - 1; j >= 0; j--) {
      if (acum[k][j] <= sum) {
        sum -= acum[k][j];
        k = par[k][j];
      }
    }
    int goUp = min(remainingL[v], 1LL * lvl[v] - lvl[k]);
    if (goUp > maxUp) {
      maxUp = goUp;
      bestSon = v;
    }
  }
  if (bestSon != -1) {
    ans--;
    remainingL[u] = remainingL[bestSon] - 1;
    remainingS[u] = remainingS[bestSon] - val[u];
  }
}
int main() {
  int n;
  scanf("%d %lld %lld", &n, &L, &S);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &val[i]);
    if (val[i] > S) {
      printf("-1\n");
      return 0;
    }
  }
  par[0][0] = 0;
  lvl[0] = 0;
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d", &par[i][0]);
    par[i][0]--;
    acum[i][0] = val[par[i][0]];
    son[par[i][0]].push_back(i);
    lvl[i] = lvl[par[i][0]] + 1;
  }
  for (int j = 1; j < 20; j++) {
    for (int i = 0; i < n; i++) {
      par[i][j] = par[par[i][j - 1]][j - 1];
      acum[i][j] = acum[par[i][j - 1]][j - 1] + acum[i][j - 1];
    }
  }
  dfs(0);
  printf("%d\n", ans);
}
