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;
int n, l, p[100010], nl[100010], vp[100010], depth[100010];
int out[100010];
bool visited[100010];
long long w[100010], ns[100010], s, wsum[100010];
int lift[100010][20], h[100010];
vector<int> children[100010];
void dfs(int u, long long p, int d) {
  wsum[u] = p;
  depth[u] = d;
  int child_count = children[u].size(), v;
  for (int i = 0; i < child_count; i++) {
    v = children[u][i];
    dfs(v, p + w[u], d + 1);
  }
}
int find_height(int u, long long s, int p) {
  int sh = 0, eh = min(depth[u], l - nl[u]), mh;
  int v;
  while (sh < eh) {
    mh = (sh + eh + 1) / 2;
    v = u;
    for (int i = 20 - 1; i >= 0; i--) {
      if ((mh >> i) & 1) {
        v = lift[v][i];
      }
    }
    if (wsum[u] - wsum[v] <= s) {
      sh = mh;
    } else {
      eh = mh - 1;
    }
  }
  return sh;
}
void topo(int u) {
  visited[u] = true;
  out[p[u]]--;
  int child_count = children[u].size(), v, mh = 0, bv = 0, vpsum = 0;
  for (int i = 0; i < child_count; i++) {
    v = children[u][i];
    if (h[v] > mh) {
      mh = h[v];
      bv = v;
    }
    vpsum += vp[v];
  }
  if (mh > 0) {
    vp[u] = vpsum;
    ns[u] = w[u] + ns[bv];
    nl[u] = nl[bv] + 1;
  } else {
    vp[u] = vpsum + 1;
    ns[u] = w[u];
    nl[u] = 1;
  }
  if (u == 1) {
    return;
  } else {
    h[u] = find_height(u, s - ns[u], p[u]);
  }
  if (out[p[u]] == 0) {
    topo(p[u]);
  }
}
int main() {
  scanf("%d %d %lld", &n, &l, &s);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    out[i] = 0;
  }
  for (int i = 2; i <= n; i++) {
    scanf("%d", &p[i]);
    out[p[i]]++;
    children[p[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (w[i] > s) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < 20; i++) {
    lift[1][i] = 0;
    lift[0][i] = 0;
  }
  for (int i = 2; i <= n; i++) {
    lift[i][0] = p[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < 20; j++) {
      lift[i][j] = lift[lift[i][j - 1]][j - 1];
    }
  }
  dfs(1, 0, 0);
  for (int i = n; i >= 1; i--) {
    if (!visited[i]) {
      topo(i);
    }
  }
  printf("%d\n", vp[1]);
  return 0;
}
