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 int desll[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int mod = 1e9 + 7;
const int maxn = 1e5 + 1;
const int maxm = 1e5;
const double eps = 1e-8;
const int csize = 2;
int n, k, m, ar[maxn];
int f[maxn], asfdasdasdfasdfasfdfasfsdfasfdas[maxn][18], v[maxn], in[maxn],
    mark;
long long ss[maxn][18];
bool ma[maxn], que[maxn];
priority_queue<pair<int, int> > qu;
vector<int> ve[maxn];
int dep[maxn];
void dfs(int u, int pre) {
  if (pre >= 0) dep[u] = dep[pre] + 1;
  for (int i = 0; i < ve[u].size(); i++) {
    int v = ve[u][i];
    if (v == pre) continue;
    dfs(v, u);
  }
}
int main() {
  int l;
  long long s, mx = 0;
  f[1] = 0;
  scanf("%d%d%I64d", &n, &l, &s);
  memset(in, 0, sizeof(in));
  memset(asfdasdasdfasdfasfdfasfsdfasfdas, 0,
         sizeof(asfdasdasdfasdfasfdfasfsdfasfdas));
  memset(ma, 0, sizeof(ma));
  memset(que, 0, sizeof(que));
  for (int i = 1; i <= n; i++) scanf("%d", &ar[i]), mx = max(mx, 1LL * ar[i]);
  for (int i = 2; i <= n; i++)
    scanf("%d", &f[i]), in[f[i]]++, ve[f[i]].push_back(i);
  dep[1] = 1;
  dfs(1, -1);
  if (mx > s) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    asfdasdasdfasdfasfdfasfsdfasfdas[i][0] = f[i];
    ss[i][0] = ar[f[i]];
  }
  for (int j = 1; j < 18; j++) {
    for (int i = 1; i <= n; i++) {
      if (i + (1 << j) > n) break;
      asfdasdasdfasdfasfdfasfsdfasfdas[i][j] = asfdasdasdfasdfasfdfasfsdfasfdas
          [asfdasdasdfasdfasfdfasfsdfasfdas[i][j - 1]][j - 1];
      ss[i][j] =
          ss[i][j - 1] + ss[asfdasdasdfasdfasfdfasfsdfasfdas[i][j - 1]][j - 1];
    }
  }
  for (int i = 1; i <= n; i++) {
    long long lmid = l - 1, smid = s - ar[i];
    int ins = 17, x = i;
    while (ins >= 0) {
      if (asfdasdasdfasdfasfdfasfsdfasfdas[x][ins] == 0)
        ins--;
      else if ((1 << ins) > lmid)
        ins--;
      else if (ss[x][ins] <= smid) {
        smid -= ss[x][ins];
        lmid -= (1 << ins);
        x = asfdasdasdfasdfasfdfasfsdfasfdas[x][ins];
      } else
        ins--;
    }
    v[i] = x;
  }
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) {
      qu.push(make_pair(dep[i], i));
      que[i] = 1;
    }
  }
  int ans = 0;
  ma[0] = 1;
  while (qu.size()) {
    int u = qu.top().second;
    qu.pop();
    if (ma[u]) continue;
    mark = v[u];
    int pre = u;
    while (dep[pre] > dep[mark]) ma[pre] = 1, pre = f[pre];
    ma[mark] = 1;
    ans++;
    if (!que[f[mark]]) {
      qu.push(make_pair(dep[f[mark]], f[mark]));
      que[f[mark]] = 1;
    }
  }
  printf("%d\n", ans);
  return 0;
}
