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 inf = 0x3f3f3f3f;
const long double eps = 1e-9;
long long mod = 998244353;
const long long N = 1e6 + 5;
const long long M = 1e5 + 5;
long long n, L, S;
long long pa[M][25];
long long w[M];
long long fa[M];
long long ans = 0;
long long far[M];
long long sum[M];
long long top[M];
long long dep[M];
struct edge {
  long long to, nx, w;
} e[M << 1];
long long fi[M];
long long ce = 1;
;
inline void add(long long u, long long v, long long w) {
  e[++ce] = (edge){v, fi[u], w};
  fi[u] = ce;
};
void wk(long long u) {
  for (long long i = 1; i <= 20; i++) pa[u][i] = pa[pa[u][i - 1]][i - 1];
  long long rem = L;
  long long x = u;
  for (long long i = 20; i >= 0; i--) {
    long long fa = pa[x][i];
    if (!fa || (1 << i) + 1 > rem) continue;
    if (sum[u] - sum[fa] + w[fa] > S) continue;
    rem -= (1 << i);
    top[u] = fa;
    x = fa;
  }
}
void dfs1(long long u, long long f) {
  dep[u] = dep[f] + 1;
  pa[u][0] = f;
  top[u] = u;
  sum[u] = w[u] + sum[f];
  wk(u);
  for (long long i = fi[u]; ~i; i = e[i].nx) {
    long long v = e[i].to;
    dfs1(v, u);
  }
}
void dfs2(long long u) {
  long long mx = -1;
  for (long long i = fi[u]; ~i; i = e[i].nx) {
    long long v = e[i].to;
    dfs2(v);
    if (far[v] == v) continue;
    if (mx == -1 || dep[far[v] < mx]) mx = far[v];
  }
  if (~mx) {
    far[u] = mx;
  } else {
    ans++;
    far[u] = top[u];
  }
}
signed main() {
  memset(fi, -1, sizeof(fi));
  scanf("%lld", &n), scanf("%lld", &L), scanf("%lld", &S);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    if (w[i] > S) {
      cout << -1;
      return 0;
    }
  }
  for (long long i = 2; i <= n; i++) {
    scanf("%lld", &fa[i]);
    add(fa[i], i, 0);
  }
  dfs1(1, 0);
  dfs2(1);
  cout << ans;
}
