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 infinity = 2000000999 / 2;
const long long int inff = 4000000000000000999;
template <class T>
T get() {
  T a;
  cin >> a;
  return a;
}
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U>& par) {
  out << "[" << par.first << ";" << par.second << "]";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const set<T>& cont) {
  out << "{";
  for (const auto& x : cont) out << x << ", ";
  out << "}";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, const map<T, U>& cont) {
  out << "{";
  for (const auto& x : cont) out << x << ", ";
  out << "}";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) out << " ";
    out << v[i];
  }
  out << endl;
  return out;
}
bool ccw(complex<long double> p, complex<long double> a,
         complex<long double> b) {
  if ((conj(a - p) * (b - p)).imag() <= 0)
    return false;
  else
    return true;
}
bool dfs_dasa(int v, int f, int h, int ms, vector<int>& s,
              vector<vector<int> >& hrany, vector<int>& maxi,
              vector<int>& res) {
  if (h % 2) {
    maxi[v] = infinity;
  } else {
    if (s[v] < ms) return false;
    maxi[v] = s[v] - ms;
  }
  for (int w : hrany[v]) {
    if (w == f) continue;
    if (!dfs_dasa(w, v, h + 1, max(ms, s[v]), s, hrany, maxi, res))
      return false;
    if (h % 2) {
      maxi[v] = min(maxi[v], maxi[w]);
    }
  }
  if (h % 2) {
    for (int w : hrany[v]) {
      if (w == f) continue;
      res[w] = maxi[w] - maxi[v];
    }
    res[v] = (maxi[v] != infinity ? maxi[v] : 0);
  }
  return true;
}
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  int n = get<int>();
  vector<vector<int> > hrany(n);
  for (int i = 0; i < (int)n - 1; i++) {
    int p = get<int>() - 1;
    hrany[p].push_back(i + 1);
  }
  vector<int> s;
  for (int i = 0; i < (int)n; i++) s.push_back(get<int>());
  vector<int> maxi(n), res(n, s[0]);
  if (dfs_dasa(0, -1, 0, 0, s, hrany, maxi, res)) {
    long long int sum = 0;
    if (0) cout << res;
    for (int i = 0; i < (int)n; i++) sum += res[i];
    cout << sum << endl;
  } else
    cout << -1 << endl;
}
