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 Mod = 1e9 + 7;
long long bfs(pair<vector<int>, pair<int, int> > G[], bool &b, int n) {
  set<int> s;
  queue<int> q;
  q.push(0);
  s.insert(0);
  int daddy[n];
  G[0].second.second = G[0].second.first;
  while (q.size() != 0) {
    int u = q.front();
    q.pop();
    if (G[u].second.first != -1) {
      for (int i = 0; i < G[u].first.size(); i++) {
        if (s.find(G[u].first[i]) == s.end()) {
          daddy[G[u].first[i]] = u;
          q.push(G[u].first[i]);
          s.insert(G[u].first[i]);
        }
      }
      continue;
    }
    int father = G[daddy[u]].second.first;
    int children = Mod;
    for (int i = 0; i < G[u].first.size(); i++) {
      if (s.find(G[u].first[i]) == s.end()) {
        if (G[G[u].first[i]].second.first < father) b = false;
        if (G[G[u].first[i]].second.first < children)
          children = G[G[u].first[i]].second.first;
      }
    }
    if (children == Mod) {
      G[u].second.second = 0;
      continue;
    }
    G[u].second.second = children - father;
    for (int i = 0; i < G[u].first.size(); i++) {
      if (s.find(G[u].first[i]) == s.end()) {
        daddy[G[u].first[i]] = u;
        q.push(G[u].first[i]);
        s.insert(G[u].first[i]);
        G[G[u].first[i]].second.second =
            G[G[u].first[i]].second.first - children;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += G[i].second.second;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  pair<vector<int>, pair<int, int> > G[n];
  int a, no;
  for (int i = 1; i < n; i++) {
    cin >> a;
    a--;
    G[a].first.push_back(i);
    G[i].first.push_back(a);
  }
  for (int i = 0; i < n; i++) {
    cin >> no;
    G[i].second.first = no;
  }
  bool b = true;
  long long ans = bfs(G, b, n);
  if (b)
    cout << ans << endl;
  else
    cout << -1 << endl;
  return 0;
}
