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;
template <class x>
vector<typename x::value_type> tovec(const x& a) {
  return vector<typename x::value_type>(a.begin(), a.end());
}
template <class u>
vector<typename iterator_traits<u>::value_type> tovec(u a, u b) {
  return vector<typename iterator_traits<u>::value_type>(a, b);
}
template <typename x>
ostream& operator<<(ostream& cout, const vector<x>& a) {
  cout << "[";
  if (((int)a.size())) cout << a[0];
  for (int i = 0; i++ < ((int)a.size()) - 1;) cout << ", " << a[i];
  return cout << "]";
}
template <typename u, typename v>
ostream& operator<<(ostream& cout, const pair<u, v>& a) {
  return cout << "(" << a.first << ", " << a.second << ")";
}
int n;
int p[101010], h[101010];
int s[101010];
vector<int> gr[101010];
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  h[1] = 1;
  p[1] = 1;
  for (int i = 2; i <= n; ++i) {
    cin >> p[i];
    h[i] = h[p[i]] + 1;
    gr[p[i]].push_back(i);
  }
  for (int i = 0; i++ < n;) {
    cin >> s[i];
  }
  long long ans = s[1];
  for (int u = 0; u++ < n;) {
    if (s[u] != -1) continue;
    if (((int)gr[u].size()) == 0) continue;
    int ms = s[gr[u][0]];
    long long sums = 0;
    for (auto v : gr[u]) {
      ms = min(ms, s[v]);
      sums += s[v];
    }
    if (ms < s[p[u]]) {
      cout << -1;
      return 0;
    }
    ans += ms - s[p[u]] + sums - ms * ((int)gr[u].size());
  }
  cout << ans;
  return 0;
}
