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 w[1000000];
int lev[1000000];
int pa[100100][20];
long long dist[1000000];
long long L, S;
vector<int> adj[1000000];
void dfs(int v) {
  for (auto u : adj[v]) {
    lev[u] = lev[v] + 1;
    dist[u] = dist[v] + w[u];
    dfs(u);
  }
}
long long bS[1000001];
long long bL[1000001];
bool ok = 1;
int ret = 0;
int getSol(int v, long long l, long long s) {
  int x = v;
  for (int i = 19; i >= 0; i--) {
    int u = pa[v][i];
    if (S - s >= dist[x] - w[x] - dist[u] + w[u]) v = u;
  }
  return max(lev[v] + 0LL, lev[x] - (L - l));
}
void sol(int v) {
  int best = lev[v] + 1;
  bL[v] = 0;
  bS[v] = 0;
  if (w[v] > S) ok = 0;
  for (auto u : adj[v]) {
    sol(u);
    int cur = getSol(u, bL[u], bS[u]);
    if (cur < best) best = cur, bS[v] = bS[u], bL[v] = bL[u];
  }
  if (best == lev[v] + 1) ret++;
  bS[v] += w[v];
  bL[v]++;
  if (ok) assert(bS[v] <= S && bL[v] <= L);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n >> L >> S;
  for (int i = 1; i <= n; i++) cin >> w[i];
  for (int i = 2; i <= n; i++) {
    cin >> pa[i][0];
    adj[pa[i][0]].push_back(i);
  }
  dist[1] = w[1];
  dfs(1);
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++) pa[j][i] = pa[pa[j][i - 1]][i - 1];
  sol(1);
  if (!ok) ret = -1;
  cout << ret << endl;
}
