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 = 1e5 + 10;
int nodeN = 0, maxSze = 0, tot = 0;
long long maxSum = 0;
int head[maxN] = {}, fa[maxN][20] = {}, far[maxN] = {}, dep[maxN] = {},
    top[maxN] = {};
long long val[maxN] = {}, toR[maxN] = {};
struct list {
  int to, next;
} edge[maxN << 1];
inline void addedge(int nu, int nv) {
  edge[++tot].to = nv;
  edge[tot].next = head[nu];
  head[nu] = tot;
}
void init() {
  scanf("%d%d%I64d", &nodeN, &maxSze, &maxSum);
  for (int nodeI = 1; nodeI <= nodeN; ++nodeI) {
    scanf("%I64d", &val[nodeI]);
    if (val[nodeI] > maxSum) {
      printf("-1\n");
      exit(0);
    }
  }
  for (int nodeI = 2; nodeI <= nodeN; ++nodeI) {
    int par;
    scanf("%d", &par);
    addedge(nodeI, par);
    addedge(par, nodeI);
  }
}
void dfs1(int now, int from) {
  fa[now][0] = from;
  toR[now] = toR[from] + val[now];
  dep[now] = dep[from] + 1;
  for (int upI = 1; upI <= 18; ++upI)
    fa[now][upI] = fa[fa[now][upI - 1]][upI - 1];
  far[now] = now;
  for (int upI = 18; upI >= 0; --upI) {
    if (fa[now][upI] && toR[now] - toR[fa[fa[far[now]][upI]][0]] <= maxSum &&
        dep[now] - dep[fa[fa[far[now]][upI]][0]] <= maxSze) {
      far[now] = fa[far[now]][upI];
    }
  }
  for (int pt = head[now]; pt; pt = edge[pt].next) {
    int to = edge[pt].to;
    if (to != from) {
      dfs1(to, now);
    }
  }
}
int dfs2(int now, int from) {
  int best = -1, ans = 0;
  for (int pt = head[now]; pt; pt = edge[pt].next) {
    int to = edge[pt].to;
    if (to != from) {
      ans += dfs2(to, now);
      if (top[to] == to) continue;
      if (best == -1 || dep[top[to]] < dep[best]) {
        best = top[to];
      }
    }
  }
  if (best == -1) {
    best = far[now];
    ++ans;
  }
  top[now] = best;
  return ans;
}
int main() {
  init();
  dfs1(1, 0);
  printf("%d\n", dfs2(1, 0));
  return 0;
}
