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 <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '[';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << ']';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '{' << p.first << ", " << p.second << '}';
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &s) {
  os << '{';
  string sep;
  for (const auto &x : s) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const map<A, B> &mp) {
  os << '[';
  string sep;
  for (const auto &x : mp) os << sep << x, sep = ", ";
  return os << ']';
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &s) {
  os << '{';
  string sep;
  for (const auto &x : s) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const long long nax = 1e5 + 5;
vector<vector<long long>> adj(nax);
vector<long long> s(nax);
vector<long long> mns(nax);
long long ans = 0;
long long n;
void dfs(long long node, long long par, long long level, long long sum) {
  long long hehe = 0;
  if (level & 1) {
    ans += abs(sum - s[node]);
    hehe = abs(sum - s[node]);
  } else {
    long long mn = mns[node];
    if (mn != 1e18) {
      hehe = abs(mn - sum);
      ans += abs(mn - sum);
    }
  }
  for (auto nei : adj[node]) {
    if (nei != par) {
      dfs(nei, node, level + 1, sum + hehe);
    }
  }
}
bool check = false;
long long dfs2(long long node, long long par) {
  long long ans = 1e18;
  for (auto nei : adj[node]) {
    if (nei != par) {
      ans = min(ans, dfs2(nei, node));
    }
  }
  if ((long long)(adj[node]).size() == 1 && node != 1) {
    return mns[node] = s[node];
  }
  mns[node] = ans;
  if (s[node] > mns[node] && s[node] != 1e18) check = true;
  return min(ans, s[node]);
}
void solve() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long x;
    cin >> x;
    adj[i + 2].push_back(x);
    adj[x].push_back(i + 2);
  }
  for (long long i = 1; i <= n; i++) {
    cin >> s[i];
    s[i] = (s[i] == -1 ? 1e18 : s[i]);
  }
  ans = 0;
  dfs2(1, 1);
  dfs(1, 1, 1, 0);
  if (check) {
    cout << -1 << '\n';
    return;
  }
  cout << ans << '\n';
}
int32_t main() {
  solve();
  return 0;
}
