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 thw, thl, dp[300002][32], a[300002], c[300002], out = 0, par[300002],
                                                          lev[300002];
pair<long long, long long> b[300002];
vector<long long> graph[300002];
void dfs0(long long u, long long p, long long level) {
  lev[u] = level;
  c[u] = a[u];
  if (p != -1) c[u] += c[p];
  for (long long j = 0; j < graph[u].size(); j++) {
    long long v = graph[u][j];
    dfs0(v, u, level + 1);
    par[v] = u;
  }
}
void create(long long n) {
  long long i, j;
  for (i = 1; i <= n; i++)
    for (j = 0; j < 31; j++) dp[i][j] = -1;
  for (i = 1; i <= n; i++) {
    dp[i][0] = par[i];
  }
  for (j = 1; j < 31; j++)
    for (i = 1; i <= n; i++) {
      if (dp[i][j - 1] != -1) {
        dp[i][j] = dp[dp[i][j - 1]][j - 1];
      }
    }
}
long long cal(long long u, long long level) {
  for (long long i = 28; i >= 0; i--) {
    if (level - (1 << i) >= 0) {
      level -= 1 << i;
      u = dp[u][i];
    }
  }
  return u;
}
long long f(long long u, long long len, long long wt) {
  long long l = 0, r = lev[u];
  while (l != r) {
    long long mid = (l + r) / 2;
    long long v = cal(u, lev[u] - mid);
    if (wt + c[u] - c[v] + a[v] - a[u] <= thw && len + lev[u] - mid <= thl)
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
void dfs(long long u) {
  long long v, j, fl = 0;
  long long ret = -1, nd = -1;
  for (long long j = 0; j < graph[u].size(); j++) {
    v = graph[u][j];
    dfs(v);
    long long ret2;
    if (b[v].second + a[u] <= thw && b[v].first + 1 <= thl) {
      ret2 = lev[u] - f(u, b[v].first + 1, b[v].second + a[u]);
      fl = 1;
      if (ret2 > ret) {
        nd = v;
        ret = ret2;
      }
    }
  }
  if (!fl) out++;
  if (nd != -1) {
    b[u].first = b[nd].first + 1;
    b[u].second = b[nd].second + a[u];
  } else {
    b[u].first = 1;
    b[u].second = a[u];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, i;
  cin >> n >> thl >> thw;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > thw) {
      cout << -1;
      return 0;
    }
  }
  par[1] = -1;
  for (i = 2; i <= n; i++) {
    long long x;
    cin >> x;
    graph[x].push_back(i);
    par[i] = x;
  }
  dfs0(1, -1, 0);
  create(n);
  dfs(1);
  cout << out;
  return 0;
}
