Mitya has a rooted tree with n vertices indexed from 1 to n, where the root has index 1. Each vertex v initially had an integer number a_v ≥ 0 written on it. For every vertex v Mitya has computed s_v: the sum of all values written on the vertices on the path from vertex v to the root, as well as h_v — the depth of vertex v, which denotes the number of vertices on the path from vertex v to the root. Clearly, s_1=a_1 and h_1=1.

Then Mitya erased all numbers a_v, and by accident he also erased all values s_v for vertices with even depth (vertices with even h_v). Your task is to restore the values a_v for every vertex, or determine that Mitya made a mistake. In case there are multiple ways to restore the values, you're required to find one which minimizes the total sum of values a_v for all vertices in the tree.

Input

The first line contains one integer n — the number of vertices in the tree (2 ≤ n ≤ 10^5). The following line contains integers p_2, p_3, ... p_n, where p_i stands for the parent of vertex with index i in the tree (1 ≤ p_i < i). The last line contains integer values s_1, s_2, ..., s_n (-1 ≤ s_v ≤ 10^9), where erased values are replaced by -1.

Output

Output one integer — the minimum total sum of all values a_v in the original tree, or -1 if such tree does not exist.

Examples

Input


5
1 1 1 1
1 -1 -1 -1 -1


Output


1


Input


5
1 2 3 1
1 -1 2 -1 -1


Output


2


Input


3
1 2
2 -1 1


Output


-1
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000000000000000;
vector<vector<long long>> graph;
long long s[100000];
long long par[100000];
long long a[100000];
long long ans = 0;
void dfs(long long u) {
  if (s[u] == -1) {
    s[u] = INF;
    for (long long v : graph[u]) {
      dfs(v);
      s[u] = min(s[u], s[v]);
    }
  } else {
    for (long long v : graph[u]) dfs(v);
  }
}
void dfs2(long long u) {
  if (s[u] != INF) {
    if (u == 0)
      ans += s[u];
    else {
      if (s[u] < s[par[u]]) {
        cout << -1 << endl;
        exit(0);
      }
      ans += s[u] - s[par[u]];
    }
  }
  for (long long v : graph[u]) dfs2(v);
}
signed main() {
  long long n;
  cin >> n;
  graph.resize(n);
  for (long long i = 1; i != n; ++i) {
    long long p;
    scanf("%lld", &p);
    graph[p - 1].push_back(i);
    par[i] = p - 1;
  }
  for (long long i = 0; i != n; ++i) scanf("%lld", &s[i]);
  dfs(0);
  dfs2(0);
  cout << ans << endl;
  return 0;
}
