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 maxn = 100010;
template <typename Tp>
inline int getmin(Tp &x, Tp y) {
  return y < x ? x = y, 1 : 0;
}
template <typename Tp>
inline int getmax(Tp &x, Tp y) {
  return y > x ? x = y, 1 : 0;
}
template <typename Tp>
inline void read(Tp &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') f = 1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  if (f) x = -x;
}
struct data {
  int v, nxt;
} edge[maxn];
int n, l, p, head[maxn], w[maxn], f[17][maxn], dep[maxn], len[maxn], mx[maxn],
    ans[maxn];
long long s, sum[maxn];
void insert(int u, int v) {
  edge[++p] = (data){v, head[u]};
  head[u] = p;
}
void dfs1(int x) {
  dep[x] = dep[f[0][x]] + 1;
  sum[x] = sum[f[0][x]] + w[x];
  for (int i = 1; i < 17; i++) f[i][x] = f[i - 1][f[i - 1][x]];
  for (int i = head[x]; i; i = edge[i].nxt) dfs1(edge[i].v);
}
void input() {
  int u;
  read(n);
  read(l);
  read(s);
  for (int i = 1; i <= n; i++) {
    read(w[i]);
    if (w[i] > s) {
      puts("-1");
      exit(0);
    }
  }
  for (int i = 2; i <= n; i++) {
    read(f[0][i]);
    insert(f[0][i], i);
  }
  dfs1(1);
  for (int i = 2; i <= n; i++) {
    u = i;
    for (int j = 16; ~j; j--)
      if (f[j][u] && dep[i] - dep[f[j][u]] < l &&
          sum[i] - sum[f[0][f[j][u]]] <= s)
        u = f[j][u];
    len[i] = dep[i] - dep[u];
  }
}
void dfs2(int x) {
  if (x ^ 1 && !head[x]) {
    ans[x] = 1;
    mx[x] = len[x];
    return;
  }
  for (int i = head[x]; i; i = edge[i].nxt) {
    dfs2(edge[i].v);
    getmax(mx[x], mx[edge[i].v]);
    ans[x] += ans[edge[i].v];
  }
  --mx[x];
  if (mx[x] == -1) mx[x] = len[x], ++ans[x];
}
int main() {
  input();
  dfs2(1);
  printf("%d\n", ans[1]);
  return 0;
}
