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 long long int mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long int N = 1e5 + 5;
const long long int LOGN = 17;
long long int dp[LOGN][N], depth[N], sum[LOGN][N];
pair<long long int, long long int> val[N];
vector<long long int> g[N];
pair<long long int, long long int> dp1[N];
long long int n, l, s;
long long int ans = 0;
void dfs(long long int u) {
  dp1[u] = {sum[0][u], 1};
  long long int d = 1e18 + 5;
  for (long long int v : g[u]) {
    dfs(v);
    if (dp1[v].first + sum[0][u] > s || dp1[v].second + 1 > l) continue;
    long long int vv = u;
    pair<long long int, long long int> lol = dp1[v];
    for (long long int k = LOGN - 1; k >= 0; k--) {
      if (lol.first + sum[k][vv] <= s &&
          lol.second + depth[vv] - depth[dp[k][vv]] <= l) {
        lol.first += sum[k][vv];
        lol.second += dp[k][vv];
        vv = dp[k][vv];
      }
    }
    if (depth[vv] < d) {
      d = depth[vv];
      dp1[u] = {dp1[v].first + sum[0][u], dp1[v].second + 1};
    }
  }
  if (d == 1e18) ans++;
}
void solve() {
  cin >> n >> l >> s;
  for (long long int i = 1; i <= n; i++) {
    cin >> sum[0][i];
    if (sum[0][i] > s) {
      cout << -1;
      return;
    }
  }
  depth[1] = 1;
  for (long long int i = 2; i <= n; i++) {
    cin >> dp[0][i];
    depth[i] = depth[dp[0][i]] + 1;
    g[dp[0][i]].push_back(i);
  }
  for (long long int k = 1; k <= LOGN - 1; k++) {
    for (long long int i = 1; i <= n; i++) {
      dp[k][i] = dp[k - 1][dp[k - 1][i]];
      sum[k][i] = sum[k - 1][i] + sum[k - 1][dp[k - 1][i]];
    }
  }
  dfs(1);
  cout << ans;
}
signed main() {
  ios ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  clock_t clk = clock();
  long long int t = 1;
  while (t--) {
    solve();
  }
  cerr << '\n' << (double)(clock() - clk) / CLOCKS_PER_SEC;
  return 0;
}
