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>
inline void inp(T &...args) {
  ((cin >> args), ...);
}
template <typename T>
inline istream &operator>>(istream &in, vector<T> &a) {
  for (auto &x : a) in >> x;
  return in;
}
template <typename T, typename U>
inline istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
const long long int inf = 1e10;
using graph = vector<vector<long long int>>;
graph g;
long long int n, m;
vector<long long int> s, a, ms;
void clearAll(long long int N) {
  N += 2;
  g.resize(N);
  s.resize(N);
  ms.resize(N);
  a.resize(N);
  for (long long int i = (0); ((1) > 0 ? i < (N) : i > (N)); i += (1)) {
    g[i].clear();
    ms[i] = 0;
    a[i] = 0;
  }
}
void dfs1(long long int u, long long int p) {
  ms[u] = (s[u] == -1 ? inf : s[u]);
  for (long long int v : g[u]) {
    if (v != p) {
      dfs1(v, u);
      ms[u] = min(ms[u], ms[v]);
    }
  }
}
void dfs2(long long int u, long long int p, long long int sm) {
  11;
  for (long long int v : g[u]) {
    if (v != p) {
      if (ms[v] == inf) {
        a[v] = 0;
        dfs2(v, u, sm);
      } else if (sm > ms[v]) {
        cout << -1 << '\n';
        exit(0);
      } else {
        a[v] = ms[v] - sm;
        dfs2(v, u, ms[v]);
      }
      if (ms[v] != s[v] and s[v] != -1) {
        cout << -1 << '\n';
        exit(0);
      }
    }
  }
}
void solve(int &T) {
  cin >> n;
  clearAll(n);
  for (long long int i = (0); ((1) > 0 ? i < (n - 1) : i > (n - 1)); i += (1)) {
    long long int x;
    cin >> x;
    g[i + 2].emplace_back(x);
    g[x].emplace_back(i + 2);
  }
  g[0].emplace_back(1);
  for (long long int i = (0); ((1) > 0 ? i < (n) : i > (n)); i += (1)) {
    cin >> s[i + 1];
  }
  dfs1(0, -1);
  {
    dfs2(0, -1, 0);
    long long int sm = 0;
    for (long long int i = (0); ((1) > 0 ? i < (n) : i > (n)); i += (1)) {
      sm += a[i + 1];
    }
    cout << sm << '\n';
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    if (0) cerr << "Case #" << i << "\n";
    solve(i);
  }
  return 0;
}
