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 int nmax = 2000 * 1000;
const int inf = 2000 * 1000 * 1000;
const long long infl = 1000ll * 1000ll * 1000ll * 1000ll * 1000ll * 1000ll;
const int mod = 1000 * 1000 * 1000 + 7;
const long double pi = acos(-1.0);
template <typename T1, typename T2>
ostream& operator<<(ostream& cout, const pair<T1, T2>& a) {
  return (cout << "(" << a.first << "; " << a.second << ")");
}
template <typename T>
ostream& operator<<(ostream& cout, const vector<T>& a) {
  if (!a.size()) {
    return (cout << "()");
  }
  cout << "(" << a[0];
  for (int i = 1; i < (int)a.size(); i++) {
    cout << "; " << a[i];
  }
  return (cout << ")");
}
template <typename T>
ostream& operator<<(ostream& cout, const set<T>& a) {
  if (!a.size()) {
    return (cout << "{}");
  }
  auto it = a.begin();
  cout << "{" << *it;
  ++it;
  for (; it != a.end(); ++it) {
    cout << "; " << *it;
  }
  return (cout << "}");
}
long long n, p[nmax], s[nmax], ss[nmax];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> p[i + 1];
    p[i + 1]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == -1) {
      ss[i] = inf;
    } else {
      ss[i] = s[i];
    }
  }
  for (int i = 1; i < n; i++) {
    if (s[p[i]] == -1) {
      ss[p[i]] = max(ss[p[p[i]]], min(ss[p[i]], ss[i]));
    }
  }
  long long answer = ss[0];
  for (int i = 1; i < n; i++) {
    if (ss[i] == inf) {
      ss[i] = ss[p[i]];
    }
    if (ss[i] - ss[p[i]] < 0) {
      cout << -1 << endl;
      return 0;
    }
    answer += ss[i] - ss[p[i]];
  }
  cout << answer << endl;
  return 0;
}
