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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
constexpr int N = 2e5 + 5;
int n;
vector<int> gr[N];
vector<int> s;
ll bfs() {
  ll ans = 0;
  queue<pair<int, ll>> q;
  q.emplace(1, 0);
  while (q.size()) {
    auto [v, sum] = q.front();
    q.pop();
    if (s[v] == -1) {
      int val = INT32_MAX;
      for (int u : gr[v]) {
        val = min(val, s[u]);
      }
      if (val == INT32_MAX) continue;
      if (val < sum) return -1;
      ans += val - sum;
      sum = val;
      for (int u : gr[v]) {
        q.emplace(u, sum);
      }
    } else {
      if (sum > s[v]) return -1;
      ans += s[v] - sum;
      sum = s[v];
      for (int u : gr[v]) {
        q.emplace(u, sum);
      }
    }
  }
  return ans;
}
void Solve() {
  cin >> n;
  s.resize(n + 1);
  for (int i = 2, p; i <= n; i++) {
    cin >> p;
    gr[p].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  cout << bfs();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  Solve();
  return 0;
}
